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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

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

Grade:11

4 Answers

Sanchit Garg
24 Points
6 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
Pushpak
16 Points
4 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
 
Pushpak
16 Points
4 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
 
Kushagra Madhukar
askIITians Faculty 628 Points
3 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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