Question

At $27^{\circ} C$ a gas suddenly compressed such that its pressure becomes $\frac{1}{8}$ th of original pressure. The temperature of the gas will be $(\gamma=5 / 3)$

• A. $-142^{\circ} C$
• B. $ 300\,K $
• C. $327^{\circ} C$
• D. $ 420\,K $

Answer 1

Answer: (A)

When the gas is suddenly compressed heat does not find time to flow in or out.
When a system undergoes a change under the condition that no exchange of heat takes place between the system and surroundings, then such a process is called adiabatic process. In this case gas is suddenly compressed, hence it is adiabatic process, the relation between temperature ( $T$ ) and pressure $(P)$ is
$\frac{T^{\gamma}}{P^{\gamma-1}}=$ constant
where $\gamma$ is ratio of specific heats.
Given, $T_{1}=27^{\circ} C=27+273=300\, K, P_{1}=P, P_{2}=\frac{P}{8}, \gamma=\frac{5}{3}$
$ \therefore \frac{T_{1}}{T_{2}}=\left(\frac{P_{1}}{P_{2}}\right)^{\frac{\gamma-1}{\gamma}} $
$\Rightarrow \frac{T_{1}}{T_{2}}=\left(\frac{8}{1}\right)^{\frac{\frac{5-1}{3}}{5 / 3}}$
$=(8)^{0.4}$
$=2.297$
$\Rightarrow T_{2}=\frac{T_{1}}{2.297}=\frac{300}{2.297}$
$=130.6\, K \simeq 131 \,K $
$\Rightarrow T_{2}=131-273$
$=-142^{\circ} C$
Note: Gas has been suddenly compressed hence its internal energy is used in doing work against external pressure and temperature of gas falls.