A certain volume of a gas at 300 K expands adiabatically until its volume is doubled. The resultant fall in…
A certain volume of a gas at 300 K expands adiabatically until its volume is doubled. The resultant fall in temperature of the gas is nearly (The ratio of the specific heats of the gas $=1.5$ )
88 K
77 K
67 K
54 K
Solution
For adiabatic process,
$\begin{aligned}
& V_2=2 V_1, \gamma=1.5, T_1=300 \mathrm{~K} \\
& \therefore \quad \mathrm{TV}^{\gamma-1}=\text { constant } \Rightarrow \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \Rightarrow 300\left(\mathrm{~V}_1\right)^{0.5}=\mathrm{T}_2\left(2 \mathrm{~V}_1\right)^{0.5} \\
& \therefore \quad \mathrm{~T}_2=300 \times\left(\frac{1}{2}\right)^{0.5}
\end{aligned}$
$\therefore \quad$ Fall in temperature is
$\Delta \mathrm{T}=\mathrm{T}_1-\mathrm{T}_2=300\left(1-\frac{1}{\sqrt{2}}\right)=88 \mathrm{~K}$