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Grade: 11

                        

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

3 years ago

Answers : (4)

Sanchit Garg
24 Points
							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
						
2 years ago
Pushpak
16 Points
							
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
 
one year ago
Pushpak
16 Points
							
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
 
one year ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
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
3 months ago
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