Question

1 mole of $C{O}_2$ gas at 300 K is expanded under the reversible adiabatic condition such that its volume becomes 27 times.
(a) What is the final temperature?
(b) What is the work done?
Given $\gamma =1.33$ and $C_V=25.08Jmo{l}^{-1}{K}^{-1}$ for $C{O}_2$

• A. $T_2=100K,w=5.016kJ$
• B. $T_2=100K,w=-5.016kJ$
• C. $T_2=100K,w=5.016J$
• D. None of these

Answer 1

Answer: (B)

$T_2=100K,w=-5.016kJ$

Given that,

$n_1=1$
$T_1=300;V_2=27V_1$

For an adiabatic change, volume and temperature are related by the expression

$T_1{V}_1^{\gamma 1}=T_2{V}_2^{\gamma 1}$

$\left(\frac{T_1}{T_2}\right)={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$

$T_2=300{\left(\frac{1}{27}\right)}^{\frac{1}{3}}=100K$

Hence, the final temperature is 100 K.

Adiabatic condition $Q=0\Rightarrow \Delta E=W=n{C}_V(T_2-T_1)$

$W=1\times 25.08\times 200=5.016kJ/mol$