A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas by heat.

a) What is the work done on the gas?

b) What is the change in its internal energy?

a) W = - 720 J

b) ΔU = 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³ (1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q = - 390 J (heat is leaving from the system)

We know that work done by gas given as

W = P (V₂ - V₁)

W = 80 x (0.003 - 0.012) KJ

W = - 0.72 KJ

W = - 720 J (Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU = 720 - 390 J

ΔU = 330 J