Chemical Thermodynamics Question 130

Question: One mole of an ideal gas is allowed to expand reversibly and adibatically from a temperature of $ 27^{o}C $ . If the work done during the process is $ 3,kJ $ , then final temperature of the gas is $ (C_{V}=20,J/K) $ [Pb. CET 2002]

Options:

A) 100 K

B) 150 K

C) 195 K

D) 255 K

Show Answer

Answer:

Correct Answer: B

Solution:

Given number of moles =1 Initial temperature $ =27^{o}C=300K $ Work done by the system $ =3,KJ=3000K $ It will be $ (-) $ because work is done by the system. Heat capacity at constant volume $ (Cv)=20,J/k $ We know that work done $ W=-nC_{V},(T_2-T_1) $ ; $ 3000=-1\times 20(T_2-300) $

$ 3000=-20T_2+6000 $

$ 20T_2=3000 $ ; $ T_2=\frac{3000}{20}=150K $



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