Question

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of ${27}^{o}C$. If the work done during the process is 3 kJ, then the final temperature of the gas is: $(C_v=20J/K/mol)$

• A. 100 K
• B. 150 K
• C. 195 K
• D. 255 K

Answer 1

Answer: (B)

150 K

Since the gas expands adiabatically ($q=0$) so the heat is totally converted into work.

For the gas, $C_v=20J/K/mol$. 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 $300-150=150$

Hence, the correct option is $\text{B}$