Respuesta :
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:
[tex]\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)[/tex]
where,
[tex]\Delta U[/tex] = change in internal energy = ?
n = number of moles of gas = 5 moles
[tex]C_v[/tex] = heat capacity at constant volume = 2R
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]
[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta U=nC_v(T_2-T_1)[/tex]
[tex]\Delta U=(5moles)\times (2R)\times (573-298)[/tex]
[tex]\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)[/tex]
[tex]\Delta U=22863.5J=22.86kJ[/tex]
The change in internal energy of the gas is 22.86 kJ.
(b) The formula used for change in enthalpy of the gas is:
[tex]\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)[/tex]
where,
[tex]\Delta H[/tex] = change in enthalpy = ?
n = number of moles of gas = 5 moles
[tex]C_p[/tex] = heat capacity at constant pressure = 3R
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]
[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta H=nC_p(T_2-T_1)[/tex]
[tex]\Delta H=(5moles)\times (3R)\times (573-298)[/tex]
[tex]\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)[/tex]
[tex]\Delta H=34295.25J=34.29kJ[/tex]
The change in enthalpy of the gas is 34.29 kJ.
