Question

# Two samples A and B, of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that γ satisfies the equation 1 − 21−γ = (γ − 1) ln2.

Solution

## Let, Initial pressure of the gas = P1 Initial volume of the gas = V1  Final pressure of the gas= P2  Final volume of the gas = V2  Given, V2 = 2 V1, for each case. In an isothermal expansion process, work done Adiabatic work done, $W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}$ It is given that same work is done in both cases. So, In an adiabatic process, ${P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma }={P}_{1}{\left(\frac{1}{2}\right)}^{\gamma }$ From eq (1), and nRT1 = P1V1 Or (γ − 1) ln 2 = 1 − 21−γPhysicsHC Verma - IIStandard XII

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