Question

A gas sample in a piston assembly expands, doing 171 kJ of work on its surroundings at the same time that 242 kJ of heat is added to the gas. (a) What is the change in internal energy of the gas during this process? (b) Will the pressure of the gas be higher or lower when these changes are completed?

Answer 1

a) ΔU = 71 kJ

b) the pressure will be higher

Step-by-step explanation:

Step 1: Data given

A gas does 171 kJ of work on its surroundings

At the same time there is 242 kJ of heat added to the gas

Step 2: Calculate change of internal energy

Change of internal energy ΔU by the gas due to the 171 kJ work done by the system and addition of 242 kJ heat to the system.

heat energy (ΔH) is the summation of heat capacity (ΔU) and work done by a system

ΔU = Q + W

The work is done by the system on its surroundings, what means energy is lost. W will have a negative value: -171 kJ

The heat is added to the system, this means we gain energy. Q will have a positive value: 242 kJ

Total change of intern energy will be:

ΔU = Q - W

ΔU = 242 kJ - 171 kJ

ΔU = 71 kJ

ΔU = nR*ΔT

For an ideal gas n and R are constant. . Now consider PV = nRT.

P and T are directly related. Therefore, an increase in temperature will result in a higher pressure.