Thermodynamics

JEE Main level, single correct, medium

18 of 190

Question 18 of 190

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ:ΔU:ΔW\Delta Q : \Delta U : \Delta W is _______.

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Answer: (D) 7:5:27 : 5 : 2

For a diatomic gas, the number of degrees of freedom is f=5f = 5.

The molar heat capacity at constant volume is Cv=f2R=52RC_v = \dfrac{f}{2}R = \dfrac{5}{2}R.

The molar heat capacity at constant pressure is Cp=Cv+R=72RC_p = C_v + R = \dfrac{7}{2}R.

At constant pressure, the heat supplied is ΔQ=nCpΔT=n(72R)ΔT\Delta Q = nC_p\Delta T = n\left(\dfrac{7}{2}R\right)\Delta T.

The change in internal energy is ΔU=nCvΔT=n(52R)ΔT\Delta U = nC_v\Delta T = n\left(\dfrac{5}{2}R\right)\Delta T.

The work done by the gas is ΔW=nRΔT=n(22R)ΔT\Delta W = nR\Delta T = n\left(\dfrac{2}{2}R\right)\Delta T.

The ratio ΔQ:ΔU:ΔW\Delta Q : \Delta U : \Delta W is 72:52:22=7:5:2\dfrac{7}{2} : \dfrac{5}{2} : \dfrac{2}{2} = 7 : 5 : 2.

Answer: 7:5:27 : 5 : 2

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