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) 1n2.

Sol. P base 1 = Initial Pressure V base 1 = Initial Volume P base 2 = Final Pressure V base 2 = Final Volume Given, V base 2 = 2V base 1, Isothermal workdone = nRT base 1 Ln(V base 2/V base 1) Adiabatic workdone = P base 1V base 1 – P base 2V base 2/γ – 1 Given that workdone in both cases is same Hence nRT base 1 Ln (V base 2/V base 1) = P base 1V base 1 – P base 2V base 2/γ – 1 ⇒ (γ – 1) In (V base 2/V base 1) = P base 1V base 1 – P base 2V base 2/nRT base 1 ⇒ (γ – 1) In (V base 2/V base 1) = nRT base 1 – nRT base 2/nRT base 1 ⇒ (γ – 1) In 2 = T base 1 – T base 1/T base 1 …(i) [∴V base 2/2V base 1] We know TV^ γ – 1 = const. in adiabatic Process. T base 1V base 1^γ – 1 = T base 2V base 2^γ – 1, or T base 1 (V base 1)^γ – 1 = T base 2 * (2)^γ – 1 * (V base 1)^γ – 1 Or, T base 1 = 2^γ – 1 or T base 2 or T base 2 = T base 1^1-γ …(ii) From (i) & (ii) (γ – 1) In 2 = T base 1 – T base 1 * 2^1-γ/T base 1 ⇒ (γ – 1) In2 = 1 – 2^1-γ