Question

A mixture of gas expands from 0.03 m3

to 0.06 m3 at a constant pressure of 1 MPa and absorbs
84 kJ of heat during the process. The change in internal energy of the mixture is

Answer 1

To find the change in internal energy of the gas mixture, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

In this case, the pressure is constant, so the work done by the system is:

W = PΔV

where P is the constant pressure and ΔV is the change in volume of the gas. Substituting the given values, we get:

W = (1 MPa)(0.06 m3 - 0.03 m3) = 0.03 MPa·m3

The heat absorbed by the gas during the process is given as 84 kJ. Substituting these values into the first law equation, we get:

ΔU = Q - W = 84 kJ - 0.03 MPa·m3 = 84 kJ - 30 kJ = 54 kJ

Therefore, the change in internal energy of the gas mixture is 54 kJ.