Question

A sample of helium (He) gas initially at 23°C and 1.0 atm is expanded from 1.3 L to 3.1 L and simultaneously heated to 44°C. Calculate the entropy change for the process.

Answer 1

Answer : The change in entropy is, 8.65 J/K

Explanation :

Formula used :

`\Delta S=nC_v\ln ((T_2)/(T_1))+nR\ln ((V_2)/(V_1))`

where,

`\Delta S` = change in entropy

n = number of moles of gas = 1 mole

R = gas constant = 8.314 J/mol.K

`C_v` = specific heat capacity at constant volume =
`(5)/(2)R` (for monoatomic gas)

`V_1` = initial volume of gas = 1.3 L

`V_2` = final volume of gas = 3.1 L

`T_1` = initial volume of gas =
`23^oC=273+23=296K`

`T_2` = final volume of gas =
`44^oC=273+44=317K`

Now put all the given values in the above formula, we get:

`\Delta S=1* (5)/(2)R\ln ((317)/(296))+1* R\ln ((3.1)/(1.3))`

`\Delta S=1* (5)/(2)* 8.314\ln ((317)/(296))+1* 8.314\ln ((3.1)/(1.3))`

`\Delta S=8.65J/K`

Therefore, the change in entropy is, 8.65 J/K