Chapter 6, Problem 6.86P

(a)

Interpretation Introduction

Interpretation:

The initial and final volume of helium bag in early morning and mid-afternoon is to be determined.

Concept introduction:

The ideal gas equation relates volume, moles, temperature, and pressure with each other.

The ideal gas equation is as follows:

$PV=nRT$ (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$n$ is the number of moles of gases.

$R$ is the gas constant.

$T$ is the temperature.

The conversion factor to convert degree Celsius to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$ (2)

Answer to Problem 6.86P

The initial and final volume of helium bag is $23.6\text{L}$ and $24.9\text{L}$ respectively.

Explanation of Solution

Substitute $15°\text{C}$ for $T_{\text{initial}}\left(°\text{C}\right)$ in the equation (2).

$\begin{array}{c}{T}_{\text{initial}}\left(\text{K}\right)=15°\text{C}+273\\ =288\text{K}\end{array}$

Substitute $30°\text{C}$ for $T_{\text{final}}\left(°\text{C}\right)$ in the equation (2).

$\begin{array}{c}{T}_{\text{final}}\left(\text{K}\right)=30°\text{C}+273\\ =303\text{K}\end{array}$

Rearrange the equation(1) to calculate the volume.

$V=\frac{nRT}{P}$ (3)

Substitute $1\text{mol}$ for $n$, $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $1\text{atm}$ for $P$ and $288\text{K}$ for $T_{\text{initial}}$ in the equation(3).

$\begin{array}{c}{V}_{\text{initial}}=\frac{\left(1\text{mol}\right)\left(0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(288\text{K}\right)}{1\text{atm}}\\ =23.6448\text{L}\\ \approx 23.6\text{L}\end{array}$

Substitute $1\text{mol}$ for $n$, $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $1\text{atm}$ for $P$ and $303\text{K}$ for $T_{\text{final}}$ in the equation(3).

$\begin{array}{c}{V}_{\text{final}}=\frac{\left(1\text{mol}\right)\left(0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(303\text{K}\right)}{1\text{atm}}\\ =24.8736\text{L}\\ \approx 24.9\text{L}\end{array}$

Conclusion

The initial and final volume of the helium bag is $23.6\text{L}$ and $24.9\text{L}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The change in internal energy is to be determined.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy $\left(\Delta E\right)$ is the difference in the energy of the product and reactant.

Answer to Problem 6.86P

The change in internal energy is $187\text{J}$.

Explanation of Solution

The formula to calculate the change in internal energy is,

$\Delta E=\frac{3}{2}nR\left(T_{\text{final}}-T_{\text{initial}}\right)$ (4)

Substitute $1\text{mol}$ for $n$, $0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$, $303\text{K}$ for $T_{\text{final}}$ and $288\text{K}$ for $T_{\text{initial}}$ in the equation (4).

$\begin{array}{c}\Delta E=\frac{3}{2}\left(1\text{mol}\right)\left(0.0821\text{L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(303\text{K}-288\text{K}\right)\\ =187.065\text{J}\\ \approx 187\text{J}\end{array}$

Conclusion

The change in internal energy is $187\text{J}$.

(c)

Interpretation Introduction

Interpretation:

Work done by helium is to be calculated.

Concept introduction:

Work $\left(w\right)$ is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is:

$w=-P\Delta V$ (5)

Here,

$w$ is the work done by the system or on the system.

$P$ is the external pressure.

$\Delta V$ is the change in the volume.

Answer to Problem 6.86P

Work done by helium is $-1.2\times {10}^2\text{J}$.

Explanation of Solution

The initial volume of gas is $23.6448\text{L}$.

The final volume of gas is $24.8736\text{L}$.

The formula to calculate $\Delta V$ of the system is:

$\Delta V=V_{\text{Final}}-V_{\text{Initial}}$ (6)

Substitute $23.6448\text{L}$ for $V_{\text{Initial}}$ and $24.8736\text{L}$ for $V_{\text{Final}}$ in the equation (6).

$\begin{array}{c}\Delta V=16.3\text{ L}-10.5\text{ L}\\ =5.8\text{L}\end{array}$

Substitute $1\text{atm}$ for $P$ and $5.8\text{L}$ for $\Delta V$ in the equation (5).

$\begin{array}{c}w=-\left(1\text{atm}\right)\left(5.8\text{L}\right)\\ =-\left(5.8\text{atm}\cdot \text{L}\right)\left(\frac{101.3\text{J}}{\text{1}\text{atm}\cdot \text{L}}\right)\\ =-124.75\text{J}\\ \approx -1.2\times {10}^2\text{J}\end{array}$

Conclusion

Work done by helium is $-1.2\times {10}^2\text{J}$.

(d)

Interpretation Introduction

Interpretation:

Heat transferred is to be calculated.

Concept introduction:

The first law of thermodynamics gives the relation between the work, heat and internal energy. According to this law “the total energy of the system plus surrounding remains constant”. The expression to the first law of thermodynamics is:

$\Delta E=q+w$ (7)

Here,

$\Delta E$ is the internal energy of the system.

$q$ is the heat released or absorbed.

$w$ is the work done by the system or on the system.

Answer to Problem 6.86P

The heat transferred is $3.1\times {10}^2\text{J}$.

Explanation of Solution

Rearange equation (7) to calculate $q$.

$q=\Delta E-w$ (8)

Substitute $187.065\text{J}$ for $\Delta E$ and $-124.75\text{J}$ for $w$ in the equation (7).

$\begin{array}{c}q=187.065\text{J}-\left(-124.75\text{J}\right)\\ =311.815\text{J}\\ \approx 3.1\times {10}^2\text{J}\end{array}$

Conclusion

The heat transferred is $3.1\times {10}^2\text{J}$.

(e)

Interpretation Introduction

Interpretation:

$\Delta H$ of the process is to be calculated.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy $\left(\Delta E\right)$ is the difference in the energy of the product and reactant but at constant pressure, the $P\Delta V$ work gets eliminated and the change in enthalpy $\left(\Delta H\right)$ is measured.

In the case of a reaction, the change in enthalpy $\left(\Delta H\right)$ is the difference in the energy of the product and reactant. The general expression to calculate $\Delta H$ is,

$\Delta H=H_{\text{Product}}-H_{\text{Reactant}}$ (9)

Here,

$\Delta H$ is the change in enthalpy of the system.

$H_{\text{Product}}$ is the enthalpy of the products.

$H_{\text{Reactant}}$ is the enthalpy of the reactants.

The heat flow at constant pressure $\left(q_{\text{p}}\right)$ is equal to the change in enthalpy change.

Answer to Problem 6.86P

$\Delta H$ of the process is $3.1\times {10}^2\text{J}$.

Explanation of Solution

The heat transferred in the process is $3.1\times {10}^2\text{J}$.

The heat flow at constant pressure $\left(q_{\text{p}}\right)$ is equal to the change in enthalpy change. Therefore, the value of the change in the enthalpy $\left(\Delta H\right)$ is $3.1\times {10}^2\text{J}$.

Conclusion

$\Delta H$ of the process is $3.1\times {10}^2\text{J}$.

(f)

Interpretation Introduction

Interpretation:

The relationship between $\Delta H$ and $q$ of the process is to be determined.

Concept introduction:

The internal energy of a process is the summation of the kinetic energy and potential energy associated with the process. In the case of a reaction, the change in internal energy $\left(\Delta E\right)$ is the difference in the energy of the product and reactant but at constant pressure, the $P\Delta V$ work gets eliminated and the change in enthalpy $\left(\Delta H\right)$ is measured.

In the case of a reaction, the change in enthalpy $\left(\Delta H\right)$ is the difference in the energy of the product and reactant. The general expression to calculate $\Delta H$ is,

$\Delta H=H_{\text{Product}}-H_{\text{Reactant}}$ (9)

Here,

$\Delta H$ is the change in enthalpy of the system.

$H_{\text{Product}}$ is the enthalpy of the products.

$H_{\text{Reactant}}$ is the enthalpy of the reactants.

Work $\left(w\right)$ is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is:

$w=-P\Delta V$ (5)

Here,

$w$ is the work done by the system or on the system.

$P$ is the external pressure.

$\Delta V$ is the change in the volume.

The first law of thermodynamics gives the relation between the work, heat and internal energy. According to this law “the total energy of the system plus surrounding remains constant”. The expression to first law of thermodynamics is:

$\Delta E=q+w$ (7)

Here,

$\Delta E$ is the internal energy of the system.

$q$ is the heat released or absorbed.

$w$ is the work done by the system or on the system.

Answer to Problem 6.86P

The heat flow at constant pressure $\left(q_{\text{p}}\right)$ is equal to the change in enthalpy change.

Explanation of Solution

The formula to calculate $\Delta H$ is as follows:

$\Delta H=\Delta E+P\Delta V$ (10)

Substitute $-w$ for $P\Delta V$ and $q+w$ for $\Delta E$ in the equation (10).

$\begin{array}{c}\Delta H=q+w+\left(-w\right)\\ =q_p\end{array}$

Conclusion

The heat flow at constant pressure $\left(q_{\text{p}}\right)$ is equal to the change in enthalpy change.