The values of q , w , Δ U and Δ H have to be calculated at constant pressure when the temperature is raised from 25 ο C to 100 ο C . Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process. Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

Question

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Answer 1

Question

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

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Answer 2

Question

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

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Answer 3

Question

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

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Answer 4

Question

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

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Answer 5

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

Answer 6

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

Answer 7

Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant pressure when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

Answer 8

(i)

Expert Solution

Answer to Problem 2B.3AE

The value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ .

The value of $w$ at constant pressure is $−0.62355 kJ_$ .

The value of $ΔU$ at constant pressure is $10.10 kJ_$ .

Answer 9

(i)

Expert Solution

Answer 10

(i)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

$Cp/(J K−1)=20.17+0.3665(T/K)$ (1)

The temperature is raised from $25 οC$ to $100 οC$ .

The conversion of Celsius to Kelvin is done as,

$0 οC=273 K$

Therefore, the conversion of $25 οC$ to Kelvin is done as,

$25 οC=25+273 K=298 K$

Similarly, the conversion of $100 οC$ to Kelvin is done as,

$100 οC=100+273 K=373 K$

At constant pressure,

$q=ΔH=∫T1T2CpdT$ (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

$ΔH=∫T1T2CpdT=∫T1T220.17dT+∫T1T20.3665 TdT=20.17(T2−T1)+0.3665(T222−T122)$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $ΔH$ .

$ΔH=20.17(373−298)+0.3665((373)22−(298)22)=20.17×75+0.18325(139129−88804)=10734.80 J=10.73 kJ_$

Therefore, the value of value of $q$ and $ΔH$ at constant pressure is $10.73 kJ_$ respectively.

The work done for a perfect gas is calculated by the formula,

$w=−pdV=−nRdT=−(1)RdT=−RdT$ (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

$w=−8.314 J mol−1 K−1×(373 K−298 K)=−8.314 J mol−1 K−1×75 K=−623.55 J=−0.62355 kJ_$

Therefore, the work done at constant pressure is $−0.62355 kJ_$ .

The first law of thermodynamics is given by the expression,

$ΔU=q+w$ (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

$ΔU=10.73 kJ+(−0.62355 kJ)=10.73 kJ−0.62355 kJ=10.10 kJ_$

Hence, the value of $ΔU$ is $10.10 kJ_$ .

Answer 13

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

Answer 14

(ii)

Interpretation Introduction

Interpretation: The values of $q, w, ΔU$ and $ΔH$ have to be calculated at constant volume when the temperature is raised from $25 οC$ to $100 οC$ .

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state. The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter. Work is defined as the form of mechanical energy that is transferred from system to another.

Answer 15

(ii)

Expert Solution

Answer to Problem 2B.3AE

The value of $ΔU$ at constant volume is $10.1112 kJ_$ .

The value of value of $ΔH$ at constant volume is $10.73475 kJ_$ .

At constant volume, $w=0$ .

The value of $q$ at constant volume is $10.1112 kJ_$ .

Answer 16

(ii)

Expert Solution

Answer 17

(ii)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

At constant volume,

$Cp−Cv=nR$ (5)

For a perfect gas,

$n=1$

Therefore,

$Cp−Cv=RCv=Cp−R$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

$Cv=Cp−RCv=20.17+0.3665×T−8.314Cv=11.856+0.3665×T$

Now,

$Cv=(∂U∂T)v∂U=∫CvdT$ (6)

Substitute the values of heat capacity at constant volume and integrate it.

$∂U=∫T1T2(11.856+0.3665×T)dT=∫T1T211.856dT+∫T1T20.3665×TdT=11.856(T2−T1)+0.3665(T222−T122)=11.856(373 K−298 K)+0.3665((373)22−(298)22)$

Simplify the above equation,

$ΔU=11.856×75+0.18325(139129−88804)=889.2+9222.0=10111.2 J=10.1112 kJ_$

Hence, the value of $ΔU$ is $10.1112 kJ_$ .

The relation between change in enthalpy, change in internal energy, and change in volume is,

$ΔH=ΔU+Δ(pV)$ (7)

Where,

Ø $ΔH$ is the change in enthalpy.
Ø $ΔU$ is the change in internal energy.
Ø $ΔV$ is the change in volume.
Ø $p$ is the pressure.

The equation (7) is also written as,

$ΔH=ΔU+Δ(pV)ΔH=ΔU+ΔpV+pΔV$

At constant volume, $pΔV=0$ . Therefore,

$ΔH=ΔU+ΔpVΔH=ΔU+V(nRT2V−nRT1V)ΔH=ΔU+nR(T2−T1)$

Substitute the values of $ΔU$ , universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

$ΔH=10111.2 J+1×8.314 J mol−1 K−1(373 K−298 K)=10111.2 J+8.314 J mol−1 K−1×75 K=10734.75 J=10.73475 kJ_$

Hence, the value of value of $ΔH$ is $10.73475 kJ_$ .

At constant volume, $w=0$ . Now, the value of $q$ is calculated as,

$ΔU=q+0ΔU=qq=10.1112 kJ_$

Hence, the value of $q$ is $10.1112 kJ_$ .

Answer 20

Ch. 2 - Prob. 2A.1ST Ch. 2 - Prob. 2B.1ST Ch. 2 - Prob. 2B.2ST Ch. 2 - Prob. 2C.1ST Ch. 2 - Prob. 2C.2ST Ch. 2 - Prob. 2D.1ST Ch. 2 - Prob. 2D.2ST Ch. 2 - Prob. 2A.1DQ Ch. 2 - Prob. 2A.2DQ Ch. 2 - Prob. 2A.3DQ

Answer to Problem 2B.3AE

Explanation of Solution

Answer to Problem 2B.3AE

Explanation of Solution

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