A mixture of gases expands from 0.03 m³ to 0.06 m³ at a constant pressure of 1 Mpa & absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is

Question

Quizplus · Accepted Answer

The change in internal energy of the gas mixture can be calculated using the first law of thermodynamics, which states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). Given that the gas absorbs 84 kJ of heat (Q = 84 kJ) and expands at a constant pressure of 1 MPa (1000 kPa), we can calculate the work done by the system during expansion using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. The change in volume (ΔV) is 0.06 m³ - 0.03 m³ = 0.03 m³. Converting the pressure to kJ by noting that 1 kPa·m³ = 1 kJ, the work done is W = 1000 kPa * 0.03 m³ = 30 kJ. Therefore, the change in internal energy (ΔU) is ΔU = Q - W = 84 kJ - 30 kJ = 54 kJ.

A Mixture of Gases Expands from 0