Question

One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of ${27}^{\circ }C$. If the work done by the gas in the process is 3kJ, the final temperature will be equal to: $(C_V=20J/Kmol)$

Answer 1

Since the gas expands adiabatically so the heat is totally converted into work.

For the gas, $C_v=20J/K$. Thus, 20 J of heat is required for $1^{\circ }$ change in temperature of the gas.

Heat change involved during the process i.e., work done $=3kJ=3000J$.

Change in temperature $=\frac{3000}{20}K=150K$

Initial temperature $=300K$

Since the gas expands so the temperature decreases and thus final temperature is $300-150=150$