Question

Calculate the heat absorbed by the atmosphere if a 12 kg air mass is warmed from 46(F) to 84(F) under a constant pressure of 540mb. How much work is done? What is the total change in internal energy?

Answer 1

molar mass of air = 28.97 gms

givem mass of air = 12 kg

so no. of moles= m/M 12000/28.97 = 414.22 moles

Cp of air = 28.98 J/mol K

Cv of air = 20.8 J/moles k

heat absorbed = n Cp * change of temp

84 F = 302.03 K

46 F = 280.92 K

heat absorbed (Q) = 414.22 * 28.97 *(302.03-280.92)

Q = 2.53 *10^5 J

internal energy U = n Cv * change of temp

U = 414.22 * 20.8 * (302.03 -280.92)

U = 1.818 *10^5 J

Now Work Done W = Q - U

W = (2.53 - 1.81) *10^5

W = 7.2 *10^4 Joules