Question

A sample of gas absorbs 4000 kJ of heat, 

(a) if volume remains constant. What is ΔU? 

(b) Suppose that in addition to absorption of heat by the sample, the surroundings does 2000 kJ of work on the sample. What is ΔU? 

(c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU?

Answer 1

Given : 

Q = + 4000 kJ 

(since heat is absorbed)

(a) Since volume remains constant, 

ΔV = 0.

W = -P_ex(V₂ – V₁) 

= -P_exΔV = -P_ex(0) = 0 

∴ ΔU = Q + W 

= 4000 + 0 

= 4000 kJ

(b) Q = + 4000 kJ 

W = + 2000 kJ 

(Work done on the system) 

ΔU = Q + W 

= 4000 + 2000 

= 6000 kJ

(c) W = -600 kJ 

(Work of expansion)

ΔU = Q + W 

ΔU = 4000 + (-600) 

= 3400 kJ