Chapter 13.2, Problem 26PP

Interpretation Introduction

Interpretation:

The temperature in Celsius is to be calculated when pressure, volume and number of moles of the gas are given.

Concept introduction:

Ideal gas equation helps to calculate temperature of a gas at given volume and pressure. According to this law,

$\mathrm{PV}=nRT$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Answer to Problem 26PP

The temperature of the gas is $-266.09C$

Explanation of Solution

Ideal gas equation helps to calculate temperature of a gas at given volume and pressure. According to this law,

$\mathrm{PV}=nRT$

Or

$\mathrm{T}=\frac{PV}{nR}$

Where

$\mathrm{P}$ is pressure of the gas

$\mathrm{V}$ is volume of the gas

$\mathrm{n}$ is the number of moles of the gas

$\mathrm{R}$ is the ideal gas constant

$\mathrm{T}$ is the temperature of the gas

Now

Number of moles of gas is $2.49mol$

$\begin{array}{c}P=143KPa\\ =143\times 10^{3}Pa\\ =143\times 10^3\times 10^{-5}bar(1Pa=10^{-5}bar)\\ =1.43bar\end{array}$

Gas constant is $0.083d{m}^{3}bar{K}^{-1}mo{l}^{-1}$

$\begin{array}{c}V=1L\\ =1d{m}^3\end{array}$

Now substituting the values into formula,

$\begin{array}{c}T=\frac{PV}{nR}\\ =\frac{1.43\times 1}{2.49\times 0.083}\\ =6.91K\end{array}$

But result is to be reported in $C$. So now converting $C$ to Kelvin:

$T(C)=T(K)-273$

Therefore

$\begin{array}{c}T\left(C\right)=6.91-273\\ =-266.09C\end{array}$

Therefore, temperature of the gas is $-266.09C$

Conclusion

The temperature of the gas at is $-266.09C$.