On giving an equal amount of heat at constant | vedclass

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Question

(A) The heat supplied at constant volume is given by the formula: $(\Delta Q)_V = n C_V \Delta T$.Since $n = 1 \, mol$,we have $(\Delta Q)_V = C_V \De

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Monoatomic

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(A) The heat supplied at constant volume is given by the formula: $(\Delta Q)_V = n C_V \Delta T$.Since $n = 1 \, mol$,we have $(\Delta Q)_V = C_V \Delta T$.We know that the molar heat capacity at constant volume is $C_V = \frac{f}{2} R$,where $f$ is the degree of freedom.Thus,$\Delta T = \frac{(\Delta Q)_V}{C_V} = \frac{(\Delta Q)_V}{\frac{f}{2} R} = \frac{2(\Delta Q)_V}{f R}$.Since $(\Delta Q)_V$ and $R$ are constant,$\Delta T \propto \frac{1}{f}$.For a monoatomic gas,$f = 3$,and for a diatomic gas,$f = 5$.Since $f_{	ext{monoatomic}}  \Delta T_{	ext{diatomic}}$.Therefore,the rise in temperature is more for the monoatomic gas.

On giving an equal amount of heat at constant volume to $1 \, mol$ of a monoatomic and a diatomic gas,the rise in temperature $(\Delta T)$ is more for:

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