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Grade 11Physical Chemistry

one mole of an ideal gas expands reversibly and adiabatically from a temperature of 27 c if the work done during the process is 3kj then the final temperature of the gas is (Cv=20J/K)
1.100 k
2.150 k
3.195 k
4.255 k

Profile image of Mayank Bhardwaj
8 Years agoGrade 11
Answers icon

4 Answers

Profile image of Sanchit Garg
8 Years ago
Answer is B.∆U=q+w. (1)For adiabatic process, q=0. So, ∆U=W.Now, Cv=dU/dT. So, dU=Cv*dT and hence W=Cv*dT. (2)Now, W=3kJ=3000J.So, Change in temperature=dT= W/Cv = 150 K.Initial temperature=27°C=300 K.So, Final temperature= 300-150 (since,dT=150) = 150 K.Regards, Sanchit GargClass XII
Profile image of Pushpak
7 Years ago
the difference between heat change at constant pressure and that at constant volume for the combustion of 46 gram of ethanol at 27 degree Celsius
 
Profile image of Pushpak
7 Years ago
the difference between heat change at constant pressure and that at mass and volume for the combustion of 46 gram of ethanol at 27 degree celsius is
 
Profile image of Kushagra Madhukar
5 Years ago
Dear student,
Please find the solution to your problem.
 
∆U = q+w. (1)
For adiabatic process, q=0.
So, ∆U = W.
Now, Cv = dU/dT.
So, dU = Cv*dT and hence W=Cv*dT. (2)
Now, W = 3kJ = 3000J.
So, Change in temperature = dT = W/Cv = 150 K.
Initial temperature = 27°C = 300 K.
Since the gas expands so the temperature decreases and thus final temperature is 300−150 = 150
 
Thanks and regards,
Kushagra

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