Chapter 6, Problem 6.76QP

(a)

Interpretation Introduction

Interpretation: For the given value of volume and number of moles of the gas at different pressure conditions, the value of temperature at each condition is to be calculated.

Concept introduction: The gas obeying all gas laws is called ideal gas and the equation used to present the relationship between pressure, volume, temperature and number of moles is ideal gas equation.

To determine: The temperature of the gas at the given conditions.

Answer to Problem 6.76QP

Solution

The temperature of the gas is $\underset{\_}{-260.8°C}$.

Explanation of Solution

Explanation

Given

The pressure of the gas is $1.00\text{atm}$.

The given mole is $1\text{mole}$.

The given volume is $1\text{L}$.

The ideal gas equation is given as,

$PV=nRT$

Where,

• $P$ is the pressure of the gas.
• $V$ is the volume of the gas.
• $T$ is the temperature of the gas.
• $n$ is the moles of the gas.
• $R$ is the universal gas constant $0.0821\text{L}\text{}\text{atm}/\text{K}\text{mol}$.

Substitute the value of $P$, $V$, $n$ and $R$ in the above equation,

$\begin{array}{c}T=\frac{PV}{nR}\\ =\frac{1.00\text{atm}\times 1\text{L}}{1\text{mol}\times 0.0821\frac{\text{L}\text{atm}}{\text{K}\text{mol}}}\\ =12.2\text{K}\end{array}$

The conversion of $°\text{C}$ into Kelvin is done as,

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

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Subtract $273$ from this value to obtain the temperature in degree Celsius as shown below,

$\begin{array}{c}T=12.2\text{K}-273\\ =\underset{\_}{-260.8°C}\end{array}$

The temperature of the gas is $\underset{\_}{-260.8°C}$.

(b)

Interpretation Introduction

To determine: The temperature of the gas at the given conditions.

Answer to Problem 6.76QP

Solution

The temperature of the gas is $\underset{\_}{-248.6°C}$.

Explanation of Solution

Explanation

Given

The pressure of the gas is $2.00\text{atm}$.

The given mole is $1\text{mole}$.

The given volume is $1\text{L}$.

The ideal gas equation is given as,

$PV=nRT$

Where,

• $P$ is the pressure of the gas.
• $V$ is the volume of the gas.
• $T$ is the temperature of the gas.
• $n$ is the moles of the gas.
• $R$ is the universal gas constant $0.0821\text{L}\text{}\text{atm}/\text{K}\text{mol}$.

Substitute the value of $P$, $V$, $n$ and $R$ in the above equation,

$\begin{array}{c}T=\frac{PV}{nR}\\ =\frac{2.00\text{atm}\times 1\text{L}}{1\text{mol}\times 0.0821\frac{\text{L}\text{atm}}{\text{K}\text{mol}}}\\ =24.4\text{K}\end{array}$

The conversion of $°\text{C}$ into Kelvin is done as,

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

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Subtract $273$ from this value to obtain the temperature in degree Celsius as shown below,

$\begin{array}{c}T=24.4\text{K}-273\\ =\underset{\_}{-248.6°C}\end{array}$

The temperature of the gas is $\underset{\_}{-248.6°C}$.

(c)

Interpretation Introduction

To determine: The temperature of the gas at the given conditions.

Answer to Problem 6.76QP

Solution

The temperature of the gas is $\underset{\_}{-263.9°C}$.

Explanation of Solution

Explanation

Given

The pressure of the gas is $0.75\text{atm}$.

The given mole is $1\text{mole}$.

The given volume is $1\text{L}$.

The ideal gas equation is given as,

$PV=nRT$

Where,

• $P$ is the pressure of the gas.
• $V$ is the volume of the gas.
• $T$ is the temperature of the gas.
• $n$ is the moles of the gas.
• $R$ is the universal gas constant $0.0821\text{L}\text{}\text{atm}/\text{K}\text{mol}$.

Substitute the value of $P$, $V$, $n$ and $R$ in the above equation,

$\begin{array}{c}T=\frac{PV}{nR}\\ =\frac{0.75\text{atm}\times 1\text{L}}{1\text{mol}\times 0.0821\frac{\text{L}\text{atm}}{\text{K}\text{mol}}}\\ =9.1\text{K}\end{array}$

The conversion of $°\text{C}$ into Kelvin is done as,

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

Where,

• $T°\text{C}$ is the temperature in degree Celsius.

Subtract $273$ from this value to obtain the temperature in degree Celsius as shown below,

$\begin{array}{c}T=9.1\text{K}-273\\ =\underset{\_}{-263.9°C}\end{array}$

The temperature of the gas is $\underset{\_}{-263.9°C}$.

Conclusion

1. a. The temperature of the gas is $\underset{\_}{-260.8°C}$.
2. b. The temperature of the gas is $\underset{\_}{-248.6°C}$.
3. c. The temperature of the gas is $\underset{\_}{-263.9°C}$