Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:
[tex] T = \frac{PV}{nR} [/tex]
[tex] T_{i} = T_{f} [/tex]
[tex] \frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR} [/tex]
Where:
[tex]P_{i}[/tex]: is the initial pressure = 5.79 atm
[tex]P_{f}[/tex]: is the final pressure =?
[tex]V_{i}[/tex]: is the initial volume = 420 cm³
[tex]V_{f}[/tex]: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
[tex] P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm [/tex]
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:
[tex] W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J [/tex]
Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
