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. $300K$
• C. $327^{\circ }C$
• D. $420K$

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, $T_1=27^{\circ }C=27+273=300K,P_1=P,P_2=\frac{P}{8},\gamma =\frac{5}{3}$
$\therefore \frac{T_1}{T_2}={\left(\frac{P_1}{P_2}\right)}^{\frac{\gamma -1}{\gamma }}$
$\Rightarrow \frac{T_1}{T_2}={\left(\frac{8}{1}\right)}^{\frac{\frac{5-1}{3}}{5/3}}$
$=(8{)}^{0.4}$
$=2.297$
$\Rightarrow T_2=\frac{T_1}{2.297}=\frac{300}{2.297}$
$=130.6K\simeq 131K$
$\Rightarrow T_2=131-273$
$=-142^{\circ }C$
Note: Gas has been suddenly compressed hence its internal energy is used in doing work against external pressure and temperature of gas falls.