JEE Mains Chemistry Past PapershardNUMERICAL

An ideal gas, V C R , is expanded adiabatically against a constant pressure of 1 atm untill it doublJEE Mains Chemistry Past Papers Chemistry Question

Question

An ideal gas, V C R , is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is ______ K (nearest integer). [ V C is the molar heat capacity at constant volume]

Answer: .

💡 Solution & Explanation

U = q + w (q = 0) nCVT = –Pext (V2 – V1) V2 = 2V1 1 1 nRT 2nRT P P  P1 = 5, T1 = 298 P2 = 5T 298  n 2 R(T2 – T1) = – 1 1 1 nRT nRT P P        Put T1 = 298 and P2 = 5T 298  Solve and we get T2 = 274.16 K T2  274 K $//(1®

💬
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes
Ask on WhatsApp →

Practice 22,000+ questions like this

AI-adaptive practice, video lectures, and full JEE Mains Chemistry content — all in one place.

JEE Advanced · JEE Mains · NEET · IChO · AP Chemistry