Question

Calculate the internal energy and enthalpy changes resulting if air changes from an initial |state of 5°C and 10bar, where its molar volume is 2.312 x103 m?mol1, to a final state of 60°C and 1 bar. Assume also that air remains a gas for which PV/T is constant and that C 20.785 and Co 29.100 Jmol K

Answer 1

Answer : The change in internal energy and change in enthalpy of the gas is 1143.2 J/mol and 1600.5 J/mol respectively.

Explanation :

(a) The formula used for change in internal energy of the gas is:

`\Delta U=C_v\Delta T\\\\\Delta U=C_v(T_2-T_1)`

where,

`\Delta U` = change in internal energy = ?

`C_v` = heat capacity at constant volume =
`20.785J/mol.K`

`T_1` = initial temperature =
`5^oC=273+5=278K`

`T_2` = final temperature =
`60^oC=273+60=333K`

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

`\Delta U=nC_v(T_2-T_1)`

`\Delta U=(20.785J/mol.K)* (333-278)K`

`\Delta U=1143.175J/mol\approx 1143.2J/mol`

The change in internal energy of the gas is 1143.2 J/mol.

(b) The formula used for change in enthalpy of the gas is:

`\Delta H=C_p\Delta T\\\\\Delta H=C_p(T_2-T_1)`

where,

`\Delta H` = change in enthalpy = ?

`C_p` = heat capacity at constant pressure =
`29.100J/mol.K`

`T_1` = initial temperature =
`5^oC=273+5=278K`

`T_2` = final temperature =
`60^oC=273+60=333K`

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

`\Delta U=nC_v(T_2-T_1)`

`\Delta U=(29.100J/mol.K)* (333-278)K`

`\Delta U=1600.5J/mol`

The change in enthalpy of the gas is 1600.5 J/mol.