Physic Labs

Problem 2

One mole of a monatomic ideal gas, initially at absolute temperature T1T_1, is heated to T2>T1T_2>T_1 in a constant-pressure process. Let RgR_g be the gas constant; assume thermodynamic equilibrium throughout and no phase change. Define WW as work done by the gas and Q>0Q>0 as heat absorbed, so the first law is ΔU=Q−W\Delta U=Q-W. (a) Use the equation of state to find the volume change and work. (b) Calculate the internal-energy change from the molecular degrees of freedom. (c) Determine the heat transfer, check its sign, and explain how it is divided between raising internal energy and doing work.
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W=Rg(T2−T1),ΔU=32Rg(T2−T1),Q=52Rg(T2−T1)\boxed{W=R_g(T_2-T_1),\quad \Delta U=\frac32R_g(T_2-T_1),\quad Q=\frac52R_g(T_2-T_1)}
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The three boxed results obey the first law and the stated work convention. All are positive because the gas is heated at constant pressure.