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 lnv3v2\ln \frac{v_3}{v_2} is _____
two reversible processes
(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)
(Given: molar heat capacity at constant volume, Cv,mC_v,m of the gas is 52\frac{5}{2}R)

Updated On: May 9, 2024
Hide Solution
collegedunia
Verified By Collegedunia

Correct Answer: 10

Solution and Explanation

Process I is adiabatic reversible

Process II is a reversible isothermal process

Process I – (Adiabatic Reversible)

ΔUR=4502250\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×1×52R(T2900)-1800R \times \frac{1 \times 5}{2} R(T_2 - 900)

T2 = 180 K

WII = –nRT2 In = W

1×R×180lnv3v21800R-1 \times R \times 180 \ln \frac{v_3}{v_2} -1800R

lnv3v2=10\ln \frac{v_3}{v_{2}}=10

Was this answer helpful?
0
0

Top Questions on Thermodynamics

View More Questions

Questions Asked in JEE Advanced exam

View More Questions