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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

```
3 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
```
2 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
```
one year 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
```
one year ago 610 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
```
5 months ago
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