Q4. A gas at a constant pressure of 2.0 × 10 Pa takes in 250 J of heat and expands by 0.50 L. Find the work done and the change in internal energy.
ΔW = PΔV = 2.0 × 10 × 0.50 × 10 = 100 J. ΔU = 250 − 100 = 150 J.
Q5. In a process the internal energy of a gas does not change, and the gas does 120 J of work. How much heat does it take in?
ΔU = 0, so ΔQ = ΔW = 120 J.
Q6. A gas is heated in a rigid, sealed container. What is the work done, and where does the heat go?
The volume cannot change, so boundary work is zero; assume no stirring, electrical or other work, giving ΔW = 0. All the heat goes into internal energy: ΔQ = ΔU, and the temperature and pressure rise.