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:

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

$\Delta U=(20.785J/mol.K)\times (333-278)K$

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

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:

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

$\Delta U=(29.100J/mol.K)\times (333-278)K$

$\Delta U=1600.5J/mol$

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