When nitrogen gas expands from 3.00 to 6.00 L and 2.00 kJ of energy is transferred as heat to the nitrogen in a cylinder fitted with a piston at an external pressure of 2.00 atm, against this constant pressure,
What is $\Delta$U for the process?

Question

Quizplus · Accepted Answer

The work done by the gas is $ W = -P \Delta V $. Converting pressure to kJ (1 atm = 101.3 J/L), $ W = -(2.00 \, \text{atm}) \times (3.00 \, \text{L}) \times (101.3 \, \text{J/L}) = -608 \, \text{J} = -0.608 \, \text{kJ} $. Using the first law of thermodynamics, $\Delta U = Q + W = 2.00 \, \text{kJ} - 0.608 \, \text{kJ} = 1.392 \, \text{kJ} \approx 1.39 \, \text{kJ}$.

When Nitrogen Gas Expands from 3 Δ\DeltaΔ U for the Process? A) -0

When Nitrogen Gas Expands from 3 $\Delta$ U for the Process?
A) -0