Question

A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (that is, no heat can flow in or out of the system) and the piston is initially held at rest. The top part of the cylinder is evacuated and the bottom part is filled with 1.00 mole of diatomic ideal gas at temperature 332 K. After the piston is released and the system comes to equilibrium, the volume occupied by gas is halved. Find the final temperature of the gas.

Answer 1

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature $T_1=323K$

Volume $V_1=V\ and\ V_2=\frac{V}{2}$

As there is no heat transfer so this is an adiabatic process

For and adiabatic process $TV^{\gamma -1}=constant$

Here $\gamma =1.4$

So $T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}$

$T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1$

$T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K$