Question

A sample of helium (He) gas initially at 19°C and 1.0 atm is expanded from 1.4 L to 2.8 L and simultaneously heated to 35°C. Calculate the entropy change for the process.

Answer 1

Step-by-step explanation:

The given data is as follows.

`T_(1)` = (19 + 273) K = 292 K,
`P_(1)` = 1.0 atm,

`V_(1)` = 1.4 L

Therefore, moles of gas present will be calculated as follows.

n =
`(P_(1)V_(1))/(RT_(1))`

=
`(1 atm * 1.4)/(0.0821 L atm/mol K * 292)`

=
`(1.4)/(23.97)`

= 0.058 mol

Since, neon is a monoatomic ideal gas hence, expression for its constant volume heat capacity is as follows.

`C_(v) = (3)/(2)R`

As the gas is expanding to a volume (
`V_(2)`) of 2.8 L and it is heated to
`35^(o)C` or (35 + 273) K = 308 K.

So, entropy change for this reaction will be calculated as follows.

`\Delta S = nC_(v) ln ((T_(2))/(T_(1))) + nR ln ((V_(2))/(V_(1)))`

=
`n * (3)/(2)R * ln ((T_(2))/(T_(1))) + nR ln ((V_(2))/(V_(1)))`

=
`nR[1.5 ln ((T_(2))/(T_(1))) + ln ((V_(2))/(V_(1)))`

=
`0.058 * 8.314 J/mol [1.5 ln ((308)/(292)) + ln ((2.8)/(1.4))`

=
`0.4822 * [1.5 * 0.052 + 0.693]`

= 0.372 J/K

Thus, we can conclude that the entropy change for the process is 0.372 J/K.