In a system composed of nitrogen gas in a cylinder fitted with a piston, when 2.00 kJ of energy is transferred as heat at an external pressure of 2.00 atm,
The nitrogen gas expands from 2.00 to 5.00 L against this constant pressure.What is $\Delta$U for the process?

Question

Quizplus · Accepted Answer

The change in internal energy ($\Delta U$) for a process can be calculated using the first law of thermodynamics, which states $\Delta U = q + w$, where $q$ is the heat added to the system and $w$ is the work done by the system. Here, $q = +2.00$ kJ (since energy is transferred as heat to the system), and work ($w$) done by the system can be calculated using the formula $w = -P\Delta V$, where $P$ is the external pressure and $\Delta V$ is the change in volume. The negative sign indicates work done by the system (expansion).Given that $P = 2.00$ atm (which needs to be converted to kJ by using the conversion factor $1 \text{ L·atm} = 101.325 \text{ J} = 0.101325 \text{ kJ}$) and $\Delta V = 5.00 \text{ L} - 2.00 \text{ L} = 3.00 \text{ L}$, the work done by the system is $w = -2.00 \text{ atm} \times 3.00 \text{ L} \times 0.101325 \text{ kJ/L·atm} = -0.60795 \text{ kJ}$.Therefore, $\Delta U = q + w = 2.00 \text{ kJ} - 0.61 \text{ kJ} \approx +1.39 \text{ kJ}$.

In a System Composed of Nitrogen Gas in a Cylinder Δ\DeltaΔ

In a System Composed of Nitrogen Gas in a Cylinder $\Delta$