Exercises — Heat and internal energy — microscopic vs macroscopic

(a) Work $W$ — an ordered macroscopic force (piston) acts through a distance. (b) Heat $Q$ — energy crosses the boundary because of a temperature difference ($\Delta T$). (c) Internal energy $U$ — energy stored inside, a state function.

$W=0$. Work by the gas is $W=P\Delta V$; at constant volume $\Delta V=0$, so no boundary moves and no work is done. Then Tool 1 gives $Q=\Delta U$: all the heat becomes stored internal energy.

Monatomic $\Rightarrow f=3$ (three translational directions only). $U=\frac{3}{2}nRT=\frac{3}{2}(1)(8.314)(300)=3741.3\text{J}\approx 3.74\text{kJ}.$

Tool 1: $\Delta U=Q-W=500-200=300\text{J}$. Of the 500 J deposited, 200 J left again as work; 300 J stayed to raise the internal energy (and the temperature).

$\Delta T=340-290=50\text{K}$. $\Delta U=\frac{3}{2}nR\Delta T=\frac{3}{2}(2)(8.314)(50)=1247.1\text{J}.$ Constant volume $\Rightarrow W=0$. Tool 1: $Q=\Delta U+W=1247.1+0=1247.1\text{J}\approx 1.25\text{kJ}$.

$\Delta U$ depends only on $T$, so it is unchanged: $\Delta U=1247.1\text{J}$. Work at constant pressure: $W=nR\Delta T=(2)(8.314)(50)=831.4\text{J}$. Tool 1: $Q=\Delta U+W=1247.1+831.4=2078.5\text{J}$. Why $Q$ is larger: at constant pressure the gas must also push the piston out, so extra heat is needed to supply that work on top of raising $U$. That extra chunk is exactly $W=nR\Delta T$ — which is why $C_P=C_V+R$.

Isothermal ⇒ $T$ constant ⇒ $\Delta U=\frac{f}{2}nR\Delta T=0$. $W=nRT\ln \frac{V_f}{V_i}=(1)(8.314)(400)\ln \frac{6.0}{2.0}=3325.6\ln 3=3653.9\text{J}.$ Tool 1 with $\Delta U=0$: $Q=W=3653.9\text{J}\approx 3.65\text{kJ}$. Interpretation: heat pours in but internal energy never rises. Microscopically, the molecules hand their kinetic energy to the receding piston as work exactly as fast as heat replenishes it, so $⟨v^2⟩$ and thus $T$ stay fixed. This is the cleanest proof that "heat in" $\ne$ "internal energy up."

Adiabatic ⇒ $Q=0$. Tool 1: $\Delta U=Q-W=0-900=-900\text{J}$. The gas paid for the work out of its own stored energy, so $U$ dropped and it cooled. $\Delta U=\frac{3}{2}nR\Delta T\Rightarrow \Delta T=\frac{\Delta U}{\frac{3}{2}nR}=\frac{-900}{\frac{3}{2}(1)(8.314)}=\frac{-900}{12.471}=-72.17\text{K}.$ $T_f=500-72.17=427.83\text{K}.$

$\Delta U$ is a state function — it depends only on the endpoints A and B, so it is the same on both paths. Path 1: $\Delta U=Q_1-W_1=800-300=500\text{J}$. Path 2 must give the same $\Delta U=500\text{J}$: $Q_2=\Delta U+W_2=500+500=1000\text{J}.$

(a) From the parent's key bridge $\frac{1}{2}m⟨v^2⟩=\frac{3}{2}{k}_{B}T$: $⟨\text{KE}⟩=\frac{3}{2}{k}_{B}T=\frac{3}{2}(1.381\times 10^{-23})(350)=7.250\times 10^{-21}\text{J}.$ (b) $U=N⟨\text{KE}⟩=(3.0\times 10^{22})(7.250\times 10^{-21})=217.5\text{J}.$ (c) Rigid tank ⇒ $W=0$, so $Q=\Delta U$. Using $\Delta U=\frac{3}{2}N{k}_B\Delta T$ with $\Delta T=70\text{K}$: $Q=\frac{3}{2}(3.0\times 10^{22})(1.381\times 10^{-23})(70)=43.50\text{J}.$

Constant volume ⇒ $Q=\Delta U=\frac{f}{2}nR\Delta T$, with $\Delta T=50\text{K}$. Monatomic ($f=3$): $Q_{\text{mono}}=\frac{3}{2}(1)(8.314)(50)=623.55\text{J}.$ Diatomic ($f=5$): $Q_{\text{di}}=\frac{5}{2}(1)(8.314)(50)=1039.25\text{J}.$ The friend is wrong. The diatomic gas has extra "storage bins" (two rotational degrees of freedom), and by equipartition each bin absorbs $\frac{1}{2}{k}_{B}T$ per molecule. More bins ⇒ more heat needed for the same $\Delta T$. See Kinetic Theory of Gases and Degrees of Freedom and Molecular Structure.

Leg A→B (constant volume, $\Delta T=300\text{K}$): $W_{AB}=0$ (no volume change). $\Delta U_{AB}=\frac{3}{2}nR\Delta T=\frac{3}{2}(1)(8.314)(300)=3741.3\text{J}.$ $Q_{AB}=\Delta U_{AB}=3741.3\text{J}.$

Leg B→C (isothermal at 600 K, $V_C/V_B=3$): $\Delta U_{BC}=0$ (temperature constant). $W_{BC}=nRT\ln \frac{V_C}{V_B}=(1)(8.314)(600)\ln 3=4988.4\ln 3=5480.8\text{J}.$ $Q_{BC}=W_{BC}=5480.8\text{J}.$

Leg C→A (constant pressure, $T$ goes $600\to 300\text{K}$, so $\Delta T=-300\text{K}$): $W_{CA}=nR\Delta T=(1)(8.314)(-300)=-2494.2\text{J}$ (work done ON the gas as it is compressed). $\Delta U_{CA}=\frac{3}{2}nR\Delta T=\frac{3}{2}(1)(8.314)(-300)=-3741.3\text{J}.$ $Q_{CA}=\Delta U_{CA}+W_{CA}=-3741.3+(-2494.2)=-6235.5\text{J}$ (heat rejected).

Whole cycle checks: $\Delta U_{\text{cycle}}=3741.3+0+(-3741.3)=0✓$ ($U$ is a state function, and a cycle returns to the start). $W_{\text{cycle}}=0+5480.8+(-2494.2)=2986.6\text{J}.$ $Q_{\text{cycle}}=3741.3+5480.8+(-6235.5)=2986.6\text{J}.$ So $Q_{\text{cycle}}=W_{\text{cycle}}=2986.6\text{J}✓$ — over any full cycle, net heat in equals net work out because internal energy comes back to where it started.

Heat is absorbed on A→B and B→C (both $Q>0$); rejected on C→A ($Q<0$). $Q_{\text{in}}=Q_{AB}+Q_{BC}=3741.3+5480.8=9222.1\text{J}.$ $\eta =\frac{W_{\text{cycle}}}{Q_{\text{in}}}=\frac{2986.6}{9222.1}=0.3238=32.38\%.$ Of every 100 J of heat fed in, about 32 J become useful net work; the rest is dumped to the cold surroundings on the compression leg.

$\Delta U$ depends only on endpoints (state function) ::: same on every path between two states. Constant volume means ::: $W=0$, so $Q=\Delta U$. Isothermal ideal gas means ::: $\Delta U=0$, so $Q=W$. Adiabatic means ::: $Q=0$, so $\Delta U=-W$ (gas cools when it expands). Cycle efficiency is ::: net work out divided by heat absorbed, not net heat.