Respuesta :
Explanation:
(a) Formula for work done in isothermal process is as follows.
[tex]W = -2.303nRT log (\frac{V_{f}}{V_{i}})[/tex]
= [tex]-2.303 \times 1 mol \times 8.314 J/mol K \times 300 K \times log (\frac{10.0 L}{5.0 L})[/tex]
= -1729 J
And, for isothermal process
[tex]\Delta U = nC_{v} \Delta T[/tex]
= [tex]nC_{v} \times 0[/tex]
According to the first law of thermodynamics,
[tex]\Delta H_{sys} = -W[/tex]
Hence, [tex]\Delta H_{sys} = -W[/tex] = 1729 J
Also, [tex]\Delta S_{sys} = \frac{\Delta H_{sys}}{T_{sys}}[/tex]
= [tex]\frac{1729}{300}[/tex]
= 5.763 J/K
Here, [tex]\Delta H_{surr} = \Delta H_{sys}[/tex] = -1729 J
So, [tex]\Delta S_{surr} = \frac{\Delta H_{surr}}{T_{surr}}[/tex]
[tex]\frac{\Delta H_{surr}}{T_{surr}}[/tex] = [tex]\frac{-1729 J}{\Delta S_{surr}}[/tex]
[tex]\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}[/tex]
= 5.763 J/K - [tex]\frac{-1729 J}{\Delta S_{surr}}[/tex]
Hence, entropy changes for reversibly by adjusting the pressure of the surroundings to match the internal pressure of the gas is 5.763 J/K - [tex]\frac{-1729 J}{\Delta S_{surr}}[/tex].
(b) Now, formula for work done in irreversible isothermal process is as follows.
W = [tex]-P_{ext} \times \Delta V[/tex]
= [tex]-2.0 \times 10^{5} Pa \times 5 \times 10^{-3} m^{3}[/tex]
= -1000 J
For isothermal irreversible process,
[tex]\Delta U[/tex] = 0
And, [tex]\Delta H_{sys}[/tex] = -W + 1000 J
[tex]\Delta S_{sys} = \frac{\Delta H_{sys}}{T_{sys}}[/tex]
= [tex]\frac{1000 J}{300 K}[/tex]
= 3.33 J/K
[tex]\Delta H_{surr} = -\Delta H_{sys}[/tex] = -1000 J
Therefore, [tex]\Delta S_{surr} = \frac{\Delta H_{surr}}{T_{surr}} = \frac{-1000 J}{T_{surr}}[/tex]
As, [tex]\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}[/tex]
= [tex]3.33 J/K - \frac{1000 J}{T_{surr}}[/tex]
Hence, for irreversibly entropy changes freely expanding in a vacuum is [tex]3.33 J/K - \frac{1000 J}{T_{surr}}[/tex].
