Question

One mole of NH3(g) is reversibly heated from 300 K to 1300 K at 1.00 bar pressure. Calculate q, w, ∆U, ∆H and ∆S.

Data

The molar heat capacity of NH3(g) is given by the equation Cpm(T) = a0 + a1T + a2T^2 with constants a0 = 24.295, a1 = 0.03990, −7.814 × 10−6 .

Cpm is in units of J K−1 mol−1 and T is in units of K.

Hints: dq = dH = CpdT and dS = dq/T = (Cp/T)dT

Answer 1

Given that:

Cpm(T) = a0 + a1T + a2T^2

also

a0 = 24.295, a1 = 0.03990, a2 = (−7.814 × 10−6)

Thus substituting values we have....

Cpm(1300)=24.295 + 51.87 + (-13.206) = 62.959 J K−1 mol−1

and,

Cpm(300)=24.295 + 11.97 + (-0.70326) = 35.562 J K−1 mol−1

so,

since cp here is temperature dependent ....

q1 = cpdT

or dq = q1-q2

= [Cpm(1300) - Cpm(300)] x (1300 - 300)

= 27.397 J K−1 mol−1 X (1000)

= 27.397 kJ mol-1

hence, dS = dq/T

= 27.397 X 1000 /(1300)

= 21.074 J K−1 mol−1

we know that dU = nCvdT

and we also know that for polyatomic gases ... cp / cv = 3

thus we have;

Cvm(1300)= (62.959 /3 )J K−1 mol−1

and,

Cvm(300)= (35.562 / 3) J K−1 mol−1

and hence similarly du = 11.8716 kJ mol-1

thus dq-dU = w

or w = (27.397 -11.8716 )kJ mol-1

=15.525 kJ mol-1