Q164 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
A gas for which $\gamma = {5}/{3}\ \left(\gamma = {C_p}/{C_v}\right)$ is heated at constant pressure. The percentage of heat supplied used for external work is
- A$20\ \%$
- Correct answerB$40\ \%$
- C$60\ \%$
- D$80\ \%$
Solution
Heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$
Work done by the gas is given by $\Delta W = P \Delta V = n R \Delta T$
The fraction of heat used for external work is $\dfrac{\Delta W}{\Delta Q} = \dfrac{n R \Delta T}{n C_p \Delta T} = \dfrac{R}{C_p}$
Using Mayer's relation $C_p - C_v = R$ and $\gamma = \dfrac{C_p}{C_v}$, we have $C_p - \dfrac{C_p}{\gamma} = R$
$\dfrac{R}{C_p} = 1 - \dfrac{1}{\gamma} = \dfrac{\gamma - 1}{\gamma}$
Given $\gamma = \dfrac{5}{3}$, substituting this value:
$\dfrac{\Delta W}{\Delta Q} = \dfrac{\dfrac{5}{3} - 1}{\dfrac{5}{3}} = \dfrac{2}{5}$
Percentage of heat used for external work is $\dfrac{2}{5} \times 100\% = 40\%$
Answer: $40\ \%$
Work done by the gas is given by $\Delta W = P \Delta V = n R \Delta T$
The fraction of heat used for external work is $\dfrac{\Delta W}{\Delta Q} = \dfrac{n R \Delta T}{n C_p \Delta T} = \dfrac{R}{C_p}$
Using Mayer's relation $C_p - C_v = R$ and $\gamma = \dfrac{C_p}{C_v}$, we have $C_p - \dfrac{C_p}{\gamma} = R$
$\dfrac{R}{C_p} = 1 - \dfrac{1}{\gamma} = \dfrac{\gamma - 1}{\gamma}$
Given $\gamma = \dfrac{5}{3}$, substituting this value:
$\dfrac{\Delta W}{\Delta Q} = \dfrac{\dfrac{5}{3} - 1}{\dfrac{5}{3}} = \dfrac{2}{5}$
Percentage of heat used for external work is $\dfrac{2}{5} \times 100\% = 40\%$
Answer: $40\ \%$
Hence, the correct answer is option: 'B'.