Question

One mole of an ideal gas is isothermally expanded from 5.0 L to 10.0 L at 300 K. Compute the entropy changes for the system, surroundings, and the universe if the process is carried out (a) reversibly by adjusting the pressure of the surroundings to match the internal pressure of the gas, and (b) irreversibly, freely expanding in a vacuum.

Answer 1

Step-by-step explanation:

(a) Formula for work done in isothermal process is as follows.

`W = -2.303nRT log ((V_(f))/(V_(i)))`

=
`-2.303 * 1 mol * 8.314 J/mol K * 300 K * log ((10.0 L)/(5.0 L))`

= -1729 J

And, for isothermal process

`\Delta U = nC_(v) \Delta T`

=
`nC_(v) * 0`

According to the first law of thermodynamics,

`\Delta H_(sys) = -W`

Hence,
`\Delta H_(sys) = -W` = 1729 J

Also,
`\Delta S_(sys) = (\Delta H_(sys))/(T_(sys))`

=
`(1729)/(300)`

= 5.763 J/K

Here,
`\Delta H_(surr) = \Delta H_(sys)` = -1729 J

So,
`\Delta S_(surr) = (\Delta H_(surr))/(T_(surr))`

`(\Delta H_(surr))/(T_(surr))` =
`(-1729 J)/(\Delta S_(surr))`

`\Delta S_(univ) = \Delta S_(sys) + \Delta S_(surr)`

= 5.763 J/K -
`(-1729 J)/(\Delta S_(surr))`

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 -
`(-1729 J)/(\Delta S_(surr))`.

(b) Now, formula for work done in irreversible isothermal process is as follows.

W =
`-P_(ext) * \Delta V`

=
`-2.0 * 10^(5) Pa * 5 * 10^(-3) m^(3)`

= -1000 J

For isothermal irreversible process,

`\Delta U` = 0

And,
`\Delta H_(sys)` = -W + 1000 J

`\Delta S_(sys) = (\Delta H_(sys))/(T_(sys))`

=
`(1000 J)/(300 K)`

= 3.33 J/K

`\Delta H_(surr) = -\Delta H_(sys)` = -1000 J

Therefore,
`\Delta S_(surr) = (\Delta H_(surr))/(T_(surr)) = (-1000 J)/(T_(surr))`

As,
`\Delta S_(univ) = \Delta S_(sys) + \Delta S_(surr)`

=
`3.33 J/K - (1000 J)/(T_(surr))`

Hence, for irreversibly entropy changes freely expanding in a vacuum is
`3.33 J/K - (1000 J)/(T_(surr))`.