Question

Nitrous oxide (N2O) behaves as an ideal gas and has a heat capacity at constant pressure CP = 38.6 J/K∙mol. 4.2 moles of N2O initially at 298 K are heated at constant pressure until a final temperature of 358 K is reached.

(a) Calculate the enthalpy change of N2O during that process.

(b) Calculate the heat transfer Q during that process.

(c) Calculate the work W performed during that process.

(d) Calculate the change in internal energy ΔU during that process.

Answer 1

Given

Number of moles , n = 4.2

T_{​​​​​​1}= 298K

T_{​​​​​​2} = 358K

C_{​​​​​​P=} 38 .6J/K-mol

Part (a)

Enthalpy change at constant pressure for an ideal gas is given as

H = n* C_{​​​​​​p}* ( T_{​​​​​​2}​​​- T_​​1)

= 4.2* 38.6* (358-298)

H = 9727.2 J

Part ( B)

Now using ist law of thermodynamic equation  

Q = U + W

​​​​​​Work done at constant pressure W = P* (V ₂ - V_{​​​​​​ 1})

where P = pressure

For an ideal gas P* V = n* R * T holds

W = n * R * ( T2 - T1)

W= 4.2 * 8.314 * ( 358- 298)

W= 2095.128 J ............ (2)

Q = n* C^{​​​​​​}_{​​​​​​v*} (T^​​​​ ₂- T ₁) + W ........ (1)

Now calculating C_{​​​​​​v}

C_{​​​​​​P} -C_{​ v} = R

Cv= C _p - R

For an ideal gas R = 8.314 J/ K- mol

C_{​​​​ v}= 38.6 - 8.314

Cv = 30.286 J / K - mol

From equation ( 1) & (2)

Q= 4.2 * 30.286* ( 358- 298) + 2095.128

Q= 9727.2 J

Part (C)

work done performed during the process

= 2095.128 J ( as calculated above in equation 2)

Part (D)

Change in internal energy is given as

U = n * Cv ( T2 - T1)

U = 4.2 * 30.286* ( 358- 298)

U = 7632.072 J