Question

Work done by $0.1\text{mole}$ of a gas at ${27}^{\circ }\text{C}$ to double its volume at constant pressure is
[Take $R=2{\text{cal mol}}^{-1}{\text{K}}^{-1}$]

• A. $54\text{cal}$
• B. $600\text{cal}$
• C. $60\text{cal}$
• D. $546\text{cal}$

Answer 1

The correct option is C $60\text{cal}$

Given
Number of moles of gas = $0.1\text{mole}$
Initial temperature $\left(T_1\right)={27}^{\circ }\text{C}=300\text{K}$

From ideal gas equation,
$PV=nRT$
$\Rightarrow P=\frac{nRT}{V}$
$\because$ Pressure is constant
$\frac{nRT}{V}=\text{constant}\Rightarrow \frac{T}{V}=\text{constant}$
Hence, $\frac{T_1}{V_1}=\frac{T_2}{V_2}$
From the data given in the question,
$V_2=2V_1$
$\Rightarrow T_2=2T_1$
$\Rightarrow \Delta T=T_2-T_1=2T_1-T_1=300\text{K}$
Work done in an isobaric process is given by
$W=P\Delta V$
Using the ideal gas equation, we can write it as,
$W=nR\Delta T$ {using ideal gas equation}
$=0.1\times 2\times 300$
$=60\text{cal}$
Thus, option (c) is the correct answer.