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 same, the value of ln V3V2 is “X”. Find value of 32X40? (given U=internal energy, S =entropy, p = pressure, R = gas constant, V = volume, and molar heat capacity of the gas at constant volume =5R2)

(I) is reversible adiabatic expansion.ΔUR=450−2250=−1800∴ΔU=−1800R(II) is an isothermal expansion∴W1=WII nCvT2−T1=−nRT2In⁡V3V2∵w(I)=ΔU=−1800RnCv(ΔT)=−1800R1X52RT2−900=−1800RAlso, 1800 R=−RT2In⁡V3V2−1800R=−180In⁡V3V2∴In⁡V3V2=10="X"32×1040=8

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 same, the value of ln $\frac{V_{3}}{V_{2}}$ is “X”. Find value of $\frac{32 X}{40}$ ? (given U=internal energy, S =entropy, p = pressure, R = gas constant, V = volume, and molar heat capacity of the gas at constant volume $= \frac{5 R}{2})$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

(I) is reversible adiabatic expansion.
$\begin{array}{l} \frac{ΔU}{R} = 450 - 2250 = - 1800 \\ ∴ ΔU = - 1800 R \end{array}$
(II) is an isothermal expansion
$∴ W_{1} = W_{II} \begin{array}{l} nCv (T_{2} - T_{1}) = - {nRT}_{2} In \frac{V_{3}}{V_{2}} \\ ∵ w (I) = ΔU = - 1800 R \\ nCv (ΔT) = - 1800 R \\ 1 X \frac{5}{2} R (T_{2} - 900) = - 1800 R \end{array}$
Also, 1800 $R=− {RT}_{2} In \frac{V_{3}}{V_{2}}$
$\begin{array}{l} - 1800 R = - 180 In \frac{V_{3}}{V_{2}} \\ ∴ In \frac{V_{3}}{V_{2}} = 10 =" X " \\ \frac{32 \times 10}{40} = 8 \end{array}$