Question

$1$ mole of $C{O}_2$ gas at $300\text{K}$ is expanded under reversible adiabatic condition such that its volume becomes $27$ times the initial volume. Calculate the work done by the gas.
$(\text{Given}\gamma =1.33\text{and}{C}_V=25.08{\text{Jmol}}^{-1}{\text{K}}^{-1}\text{For}C{O}_2)$

• A. $10\text{kJ/mol}$
• B. $5\text{kJ/mol}$
• C. $20\text{kJ/mol}$
• D. $25\text{kJ/mol}$

Answer 1

Answer: (B)

Given, number of moles $\left(n\right)=1\text{mole}$
Intitial temperature, $\left(T_1\right)=300\text{K}$
Intial volume = $V_1$
Final Volume = $27V_1$
In an reversible adiabatic condition,
$T_1{V}_1^{(\gamma -1)}=T_2{V}_2^{(\gamma -1)}$
$\Rightarrow \left(\frac{T_1}{T_2}\right)={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$
$\Rightarrow T_2=300{\left(\frac{1}{27}\right)}^{\frac{1}{3}}=100\text{K}$
At adiabatic condition,
$q=0$
$\Rightarrow \Delta E=W=n{C}_V(T_2-T_1)$
Where, $E=\text{internal energy}$
$W=\text{Work done by the gas}$
$\therefore W=1\times 25\times \left(200\right)=5\text{kJ/mol}$