A stationary mass of gas is compressed without friction from an initial state of 0.3 m³ and 0.105 MPa to a final state of 0.15 m³ and 0.105 MPa, the pressure remaining constant during the process. There is a transfer of 40 kJ of heat from the gas during the process. How much does the internal energy of the gas change?

Concept:

The First Law of Thermodynamics for a process (state 1 to state 2) is given by-

Q1-2 = ΔU + W1-2

Q1-2 = (U2 - U1) + W1-2

where Q and W is heat and work interaction between system and surrounding.

Calculation:

Given:

Q = 40 kJ of heat from the gas [negative]

V₁ = 0.3 m3 and P1 = 0.105 MPa = 0.105 × 10³ kPa

V₂ = 0.15 m3 and P2 = 0.105 MPa = 0.105 × 103 kPa

Pressure remaining constant, therefore work-interaction will be:

W1-2 = P(V₂ - V₁)

Q1-2 = (U2 - U1) + W1-2

Q1-2 = (U2 - U1) + P(V2 - V1)

(U2 - U1) = Q1-2 - P(V2 - V1)

(U2 - U1) = -40 - [0.105 × 10³ × (0.15 - 0.3)]

(U2 - U1) = -40 - [-15.75]

(U2 - U1) = -40 + 15.75 = -24.25 kJ