One mole of a monoatomic ideal gas undergoes | vedclass

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(C) For a process,the molar heat capacity is given by $C = C_{V} + \frac{P}{n} \frac{dV}{dT}$.For a monoatomic gas,$C_{V} = \frac{3}{2} R$.The process

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$2 R$

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(C) For a process,the molar heat capacity is given by $C = C_{V} + \frac{P}{n} \frac{dV}{dT}$.For a monoatomic gas,$C_{V} = \frac{3}{2} R$.The process is a straight line in the $V-T$ diagram passing through $(V_{0}, T_{0})$ and $(2V_{0}, 3T_{0})$.The equation of the line is $V - V_{0} = \frac{2V_{0} - V_{0}}{3T_{0} - T_{0}} (T - T_{0}) = \frac{V_{0}}{2T_{0}} (T - T_{0})$.Thus,$\frac{dV}{dT} = \frac{V_{0}}{2T_{0}}$.Using the ideal gas law $PV = nRT$,we have $P = \frac{nRT}{V}$.At the point $(V_{0}, T_{0})$,$P = \frac{nRT_{0}}{V_{0}}$.Substituting these into the heat capacity formula:$C = \frac{3}{2} R + \left( \frac{nRT_{0}}{V_{0}} \right) \frac{1}{n} \left( \frac{V_{0}}{2T_{0}} \right) = \frac{3}{2} R + \frac{R}{2} = 2R$.

One mole of a monoatomic ideal gas undergoes a quasistatic process,which is depicted by a straight line joining points $(V_{0}, T_{0})$ and $(2 V_{0}, 3 T_{0})$ in a $V-T$ diagram. What is the value of the heat capacity of the gas at the point $(V_{0}, T_{0})$?

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