Question:
Starting at temperature 300 K, one mole of an ideal diatomic gas (\(\gamma = 1.4\)) is first compressed adiabatically from volume \(V_1\) to \(V_2 = \frac{V_1}{16}\). It is then allowed to expand isobarically to volume \(2V_2\). If all the processes are the quasi-static then the final temperature of the gas (in \(^o\)K) is (to the nearest integer) _____.
Starting at temperature 300 K, one mole of an ideal diatomic gas (\(\gamma = 1.4\)) is first compressed adiabatically from volume \(V_1\) to \(V_2 = \frac{V_1}{16}\). It is then allowed to expand isobarically to volume \(2V_2\). If all the processes are the quasi-static then the final temperature of the gas (in \(^o\)K) is (to the nearest integer) _____.
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Use \(TV^{\gamma-1} = \text{constant}\) for the adiabatic step, then \(V/T = \text{constant}\) at constant pressure.
Updated On: Nov 5, 2026
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Correct Answer: 1819
Solution and Explanation
Adiabatic compression: \(T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}\), so \(T_2 = 300\times 16^{0.4} = 300\times 2^{1.6} = 300\times 3.0314 = 909.4\) K.
Isobaric expansion from \(V_2\) to \(2V_2\): \(\frac{T_3}{T_2} = \frac{2V_2}{V_2} = 2\), so \(T_3 = 2\times 909.4 \approx 1818.9\) K.
To the nearest integer: 1819 K.
Answer: 1819
Isobaric expansion from \(V_2\) to \(2V_2\): \(\frac{T_3}{T_2} = \frac{2V_2}{V_2} = 2\), so \(T_3 = 2\times 909.4 \approx 1818.9\) K.
To the nearest integer: 1819 K.
Answer: 1819
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