For an ideal mono atomic gas during any process T $=$ kV, the molar heat capacity of the gas during the process is: (assume vibrational degree of freedom, to be active)

\frac{5}{2} R

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3 R

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\frac{7}{2} R

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4 R

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Solution

The correct option is A $\frac{5}{2} R$
For an ideal monoatomic gas, the enthalpy is $\frac{5}{2} R T$ .
The change in enthalpy is $\partial H = \frac{5}{2} R \partial T$ .
Hence, the molar heat capacity is ${(\frac{\partial H}{\partial T})}_{P} = \frac{5}{2} R$ .

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For an ideal mono atomic gas during any process T = kV, the molar heat capacity of the gas during the process is: (assume vibrational degree of freedom, to be active)

The correct option is A 52RFor an ideal monoatomic gas, the enthalpy is 52RT.The change in enthalpy is ∂H=52R∂T.Hence, the molar heat capacity is (∂H∂T)P=52R.

For an ideal mono atomic gas during any process T $=$ kV, the molar heat capacity of the gas during the process is: (assume vibrational degree of freedom, to be active)

The correct option is A $\frac{5}{2} R$
For an ideal monoatomic gas, the enthalpy is $\frac{5}{2} R T$ .
The change in enthalpy is $\partial H = \frac{5}{2} R \partial T$ .
Hence, the molar heat capacity is ${(\frac{\partial H}{\partial T})}_{P} = \frac{5}{2} R$ .