Thermodynamics — Medium JEE chemistry MCQ
One mole of an ideal monoatomic gas (γ = 5/3) undergoes a reversible adiabatic expansion from initial temperature 400 K and volume 2 L to final volume 16 L. The final temperature of the gas is:
- A. 50 K
- B. 100 K
- C. 200 K
- D. 25 K
Solution
For a reversible adiabatic process:
\[
TV^{\gamma-1} = \text{constant}
\]
Therefore:
\[
T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}
\]
\[
T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}
\]
For monoatomic gas, γ = 5/3, so γ - 1 = 2/3:
\[
T_2 = 400 \times \left(\frac{2}{16}\right)^{2/3} = 400 \times \left(\frac{1}{8}\right)^{2/3}
\]
\[
T_2 = 400 \times \frac{1}{8^{2/3}} = 400 \times \frac{1}{4} = 100\,\text{K}
\]
So the correct option is B.