Physic Labs

Problem 2

One mole of a monatomic ideal gas starts at (P0,V0,T0)(P_0,V_0,T_0). It is heated at constant pressure until its volume is 2V02V_0, then cooled at constant volume back to temperature T0T_0. Find the work done by the gas, the heat received by it over the full process, and its change in internal energy. Heat entering the gas is positive. Treat the gas as a closed system, apply the equation of state at each state, and use the stated sign convention to distinguish work from heat. (a) Relate temperature to volume on the first leg and specify pressure and volume at the junction. (b) Find the work on each leg and the total change in internal energy. (c) Use the first law to find net heat and explain why its sign need not match the heat on the cooling leg.
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ΔU=Q−W⇒Q=W=RT0=P0V0\Delta U=Q-W\quad\Rightarrow\quad Q=W=RT_0=P_0V_0
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With the stated convention, the internal-energy change equals heat received minus work done by the gas. The initial and final temperatures are both T_0, so the internal-energy change vanishes and net heat Q equals work W, namely RT_0=P_0V_0.