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

Correct Answer: 10

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

W_I = ∆U = -1800R

Process II – (Reversible Isothermal Process)

T₁ = 900 K

Calculation of T₂ after the reversible adiabatic process

–1800R = nC_v(T₂ – T₁)

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

T₂ = 180 K

W_II = –nRT₂ In = W

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

\(\ln \frac{v_3}{v_{2}}=10\)