A classical ideal gas is subjected to a reversible process in which its molar specific heat changes with temperature T as C(T)=CV+RTT0C(T) = C_V + R $$\frac{T}{T_0}$$C(T)=CV+RT0T. If the initial temperature and volume are T0 and V0, respectively, and the final volume is 2V0, then the final temperature is

T0ln⁡$$2\frac{T_0}{\ln 2}$$ln2T0

T01−ln⁡$$2\frac{T_0}{1 - \ln 2}1$$−ln2T0

T0(1+ln⁡2)T_0 (1 + \ln 2)T0(1+ln2)

A classical ideal gas is subjected to a reversible process in which its molar specific heat changes with temperature T as C(T)=CV+RTT0C(T) = C_V + R $$\frac{T}{T_0}$$C(T)=CV+RT0T. If the initial temperature and volume are T0 and V0, respectively, and the final volume is 2V0, then the final temperature is

Solution:From the first law of thermodynamics:Start with the general form of the first law:dQ = dU + p dVSubstitute specific heat terms:dQ = Cv dT + p dV T=T0(1+ln⁡2)T = T_0 \\left( 1 + \\ln 2 \\right)T=T0(1+ln2)

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A classical ideal gas is subjected to a reversible process in which its molar specific heat changes with temperature T as C(T)=CV+RTT0C(T) = C_V + R \

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