Question

Two moles of an ideal monoatomic gas occupies a volume V at $27{}^{\circ }C$. The gas expands adiabatically to a volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

• A. (a) 295 K (b) - 2.7 kJ
• B. (a) 189 K (b) -2.7 kJ
• C. (a) 295 K (b) 2.7 kJ
• D. (a) 189 K (b) 2.7 kJ

Answer 1

The correct option is B (a) 189 K (b) -2.7 kJ

a) We know,
$T{V}^{\gamma -1}$ = constant. (Poissons equation)
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$ .....(i)
According to the given data,
$T_1=300K,V_1=V,T_2=TK\left(assumed\right),V_2=2V$
The value of $\gamma$ for monoatomic gas = $\frac{5}{3}=1.667$
$=300\times V^{1.667-1}=T\times (2V{)}^{1.667-1}$ (using equation (i))
$=300=T\times (2{)}^{1.667-1}$
$=T=\frac{300}{1.588}$
$\Rightarrow T=188.9K$
b) $dU=n{C}_{v}dT$
​​​​​​$=2\times \frac{3R}{2}(188.9-300)$​
$=-2771.05J$
$=-2.77kJ$