Question

A certain mass of gas at NTP is expanded to three times its volume under adiabatic conditions. The resulting temperature of gas ,' will be ($7$ for gas is $ 1.40 $ )

• A. $273 \times\left(\frac{1}{3}\right)^{1.4}$
• B. $ 273\times (3)^{0.4} $
• C. $ 273\times \left( \frac{1}{3} \right)^{0.4} $
• D. $ 273\times (3)^{1.4} $

Answer 1

Answer: (C)

An adiabatic process is one in which no heat is transferred to or from working fluid.
For a gas undergoing adiabatic process $ TV^{\gamma -1} $ = constant where $T$ is temperature, $V$ the volume and $ \gamma $ the ratio of specific heat.
$\therefore \frac{T_{2}}{T_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}$
Given, $T_{1}=273\, K$
$V_{2}=3 \,V_{1}$
$\gamma=1.4$
$\therefore T_{2}=273 \times\left(\frac{V_{1}}{3 V_{1}}\right)^{1.4-1}$
$=273 \times\left(\frac{1}{3}\right)^{0.4}$