Question

5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas (b) The change in enthalpy of the gas.

Answer 1

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

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

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

where,

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

n = number of moles of gas = 5 moles

$C_v$ = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

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

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

$\Delta U=(5moles)\times (2R)\times (573-298)$

$\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)$

$\Delta U=22863.5J=22.86kJ$

The change in internal energy of the gas is 22.86 kJ.

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

$\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)$

where,

$\Delta H$ = change in enthalpy = ?

n = number of moles of gas = 5 moles

$C_p$ = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $25^oC=273+25=298K$

$T_2$ = final temperature = $300^oC=273+300=573K$

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

$\Delta H=nC_p(T_2-T_1)$

$\Delta H=(5moles)\times (3R)\times (573-298)$

$\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)$

$\Delta H=34295.25J=34.29kJ$

The change in enthalpy of the gas is 34.29 kJ.