Question

The molar heat capacity at constant pressure Cp,m of certain ideal gas was found to vary according to the expression

Cp,m = co + ciT, where co = 6.723 J K-1 mol-1 and cı = 0.1222 J K-2 mol-1 are constants peculiar to the gas. Calculate q, w, AU, and AH for a system comprising 3.0 mol of the gas undergoing the following reversible transformations:

Answer 1

ANSWER

In this exercise there is two tyope of process: a) isobaric = constant pressure and b) isochoric = constant volume. The definitions of q, w, ΔU and ΔH for each process are

Process		q	w	ΔU	ΔH
Isobaric	ΔP = 0	q = n Cp,m ΔT	w=n R ΔT	ΔU = q - w	ΔH = q
Isochoric	ΔV = 0	q = n Cv,m ΔT	w = 0	ΔU = q	ΔH = n Cp,m ΔT

We need to find the C_p.m and C_v.m values at 25 ºC and 100 ºC. In this case C_p,m is defined as

Cp, m = C +CT

Cp.m = Cum + R → Cum = Cp.m - R

• at 25 ºC (298 K):

Cp,m = 6.723 12 mol. K TW.144- mol.K2400 (298 K) = 43.1386 0 .mol.K +0.1222 -

Cv.m = 43.1386 mol.k-8,314 mol.K = 34.8246 mol. K

• at 100 ºC (373 K):

Cp,m = 6.723 12 mol.K +0.1222 mol.K? (373 K) = 52.3036 - mol.K

Cum = 52.3036 - mol.k-8.314 mol. = 43.9896 mol.K

Now we can calculate q, w, ΔU and ΔH for each process:

a) Isochoric process: from 25 ºC to 100 ºC

• Work (w)

w = nRAT = 3 mol x 8.314 (373 K – 298 K)

w = 1870,65 J

• Heat (q) and enthalpy (ΔH)

H = q = nCp,m47

H = H 373 K) - H298 K) = nCp,m (373 K7 373 K) – np.m (298 KT 298 K)

$\Delta H=q=3\: mol \left ( 52.3036\: \frac{J}{mol.K} \times 373\: K-43.1386\: \frac{J}{mol.K} \times298\: K \right )$

$\Delta H=q=\mathbf{19961.82\: J}$

• Change of Internal energy (ΔU)

$\Delta U=q-w=19961.82\: J-1870.65\: J=\mathbf{18091.17\: J}$

a) Isochoric process: from 25 ºC to 100 ºC

• Heat (q) and Change of Internal energy (ΔU)

$\Delta U=q=nC_{v,m}\Delta T$

SU = U(373 K) - U (298 K) = nCom (373 KT373 K) – nCum (298 KT 298 K

$\Delta U=q=3\: mol \left ( 43.9896\: \frac{J}{mol.K} \times 373\: K-34.8246\: \frac{J}{mol.K} \times298\: K \right )$

$\Delta U=\mathbf{18091.17\: J}$

• Enthalpy (ΔH)

H = H 373 K) - H298 K) = nCp,m (373 K7 373 K) – np.m (298 KT 298 K)

$\Delta H=3\: mol \left ( 52.3036\: \frac{J}{mol.K} \times 373\: K-43.1386\: \frac{J}{mol.K} \times298\: K \right )$

$\Delta H=\mathbf{19961.82\: J}$

CONCLUSIONS

Process	q	w	ΔU	ΔH
Isobaric	19961.82 J 19,96 kJ	1870.65 J 1.87 kJ	18091.17 J 18.09 kJ	19961.82 J 19,96 kJ
Isochoric	18091.17 J 18.09 kJ	0 J	18091.17 J 18.09 kJ	19961.82 J 19,96 kJ