Answer :
Answer:
(a). The final volume is 2.5 times the initial volume.
The work done is 1143 J.
(b). The final temperature is 207.9 K
The work done is 574 J.
Explanation:
Given that,
Sample of an ideal gas = 0.500 mol
Initial pressure = 400 kPa
Final pressure = 160 kPa
Temperature = 300 K
(a) for isothermal,
Temperature will be same.
We need to calculate the volume of gas
[tex]P_{f}V_{f}=P_{i}V_{i}[/tex]
[tex]V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}[/tex]
Put the value into the fomrula
[tex]V_{f}=(\dfrac{400}{160})V_{i}[/tex]
[tex]V_{f}=2.5 V_{i}[/tex]
We need to calculate the work done
Using equation of energy
[tex]dQ=dW[/tex]
[tex]dQ=nRTln(\dfrac{P_{in}}{P_{f}})[/tex]
[tex]dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})[/tex]
[tex]dQ=1143\ J[/tex]
(b). For adiabatic,
No transfer of heat between system and surroundings
We need to calculate the final temperature
Using formula of gas
[tex]P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}[/tex]
[tex]T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}[/tex]
Put the value into the formula
[tex]T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300[/tex]
[tex]T_{f}=207.9\ K[/tex]
We need to calculate the wok done in adiabatic
Using formula of work done
[tex]W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}[/tex]
[tex]W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}[/tex]
[tex]W=574\ J[/tex]
Hence, (a). The final volume is 2.5 times the initial volume.
The work done is 1143 J.
(b). The final temperature is 207.9 K
The work done is 574 J.