Question

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are the same, the value of $\ln \frac{v_3}{v_2}$ is _____

(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)
(Given: molar heat capacity at constant volume, $C_v,m$ of the gas is $\frac{5}{2}$R)

Answer 1

Process I is adiabatic reversible

Process II is a reversible isothermal process

Process I – (Adiabatic Reversible)

$\frac{\Delta U}{R} = 450 - 2250$

∆U = -1800 R

WI = ∆U = -1800R

Process II – (Reversible Isothermal Process)

T1 = 900 K

Calculation of T2 after the reversible adiabatic process

–1800R = nCv(T2 – T1)

$-1800R \times \frac{1 \times 5}{2} R(T_2 - 900)$

T2 = 180 K

WII = –nRT2 In = W

$-1 \times R \times 180 \ln \frac{v_3}{v_2} -1800R$

$\ln \frac{v_3}{v_{2}}=10$