Exercises — Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation)

You know $V$ and want $T$, with no pressure in sight. The relation containing only $T$ and $V$ is $T{V}^{\gamma -1}=\text{const}.$ Why this one: it links exactly the two quantities you have and want, so no substitution is needed. Reaching for $P{V}^{\gamma }$ would force you to first find pressures — extra work and extra places to slip. Answer ::: $T{V}^{\gamma -1}=\text{const}$.

$P{V}^{\gamma }$ carries the full $\gamma$; $T{V}^{\gamma -1}$ carries $\gamma$ minus one. Why the difference: swapping $P\leftrightarrow T$ using $PV=nRT$ trades one factor of $V$, so the $V$-exponent shifts by exactly $1$. Answer ::: $P{V}^{\gamma }$, $T{V}^{\gamma -1}$.

First, notation: the subscript on a derivative names what is held constant as we take it. So ${\left(\frac{dP}{dV}\right)}_T$ is "how $P$ changes with $V$ while temperature $T$ is fixed" — the slope along an isotherm. Likewise ${\left(\frac{dP}{dV}\right)}_Q$ means "…while $Q$ (the heat) is fixed at zero" — the slope along an adiabat. The subscript $Q$ therefore just labels the no-heat (adiabatic) constraint. With that, the two slopes are ${\left(\frac{dP}{dV}\right)}_T=-\frac{P}{V},{\left(\frac{dP}{dV}\right)}_Q=-\gamma \frac{P}{V}.$ Since $\gamma >1$, the adiabat is steeper by a factor $\gamma$. The figure makes this concrete: both curves pass through the same violet point, but the magenta adiabat plunges more steeply than the orange isotherm. Read off the two annotated slope arrows — the adiabat's arrow tilts down harder because, as the gas expands, it also cools, so its pressure falls faster than the isotherm where incoming heat props $T$ (and hence $P$) up.

Use $T{V}^{\gamma -1}=\text{const}$: $T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300(3{)}^{0.4}.$ $3^{0.4}=e^{0.4\ln 3}=e^{0.4394}=1.5518$. So $T_2=300\times 1.5518\approx 465.5\text{K}.$ Sanity: compressing squeezes work into the gas with no heat escaping, so it must heat up — $465\text{K}>300\text{K}$. ✔️ Answer ::: $T_2\approx 466\text{K}$.

Use $P{V}^{\gamma }=\text{const}$: $P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=10(1/2{)}^{1.4}.$ $(1/2{)}^{1.4}=e^{-1.4\ln 2}=e^{-0.9704}=0.3789$. So $P_2=10\times 0.3789\approx 3.79\text{atm}.$ Sanity: expansion drops pressure, and because it also cools the gas the drop is steeper than isothermal (which would give $5\text{atm}$). ✔️ Answer ::: $P_2\approx 3.79\text{atm}$.

We have $P$ and $T$, no volumes — so use $T^{\gamma }{P}^{1-\gamma }=\text{const}$, i.e. $T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{(\gamma -1)/\gamma }$. Here $\frac{\gamma -1}{\gamma }=\frac{2/3}{5/3}=0.4$, and $\frac{P_2}{P_1}=\frac{1}{3}$. $T_2=400(1/3{)}^{0.4}=400e^{-0.4\ln 3}=400e^{-0.4394}=400\times 0.6444=257.8\text{K}.$ Sanity: pressure fell → gas expanded → it cooled: $258\text{K}<400\text{K}$. ✔️ Answer ::: $T_2\approx 258\text{K}$.

Adiabatic means $Q=0$, so the first law $0=\Delta U+W$ gives $W=-\Delta U=n{C}_V(T_1-T_2)$. With $C_V=\frac{5}{2}R=2.5\times 8.314=20.785{\text{J mol}}^{-1}{\text{K}}^{-1}$: $W=2\times 20.785\times (500-350)=2\times 20.785\times 150=6235.5\text{J}.$ Why positive: the gas expands, so it does work on its surroundings; that energy is drawn from its own internal store, which is why it cooled. Positive $W$ and falling $T$ are the same fact seen twice. ✔️ Answer ::: $W\approx 6236\text{J}$.

Isotherm: ${\left(\frac{dP}{dV}\right)}_T=-\frac{P}{V}=-\frac{4}{10}=-0.4\text{atm/L}$. Adiabat: ${\left(\frac{dP}{dV}\right)}_Q=-\gamma \frac{P}{V}=-1.4\times 0.4=-0.56\text{atm/L}$. Ratio $=\frac{-0.56}{-0.4}=1.4=\gamma$. Why: both slopes share the factor $P/V$; the adiabat carries an extra $\gamma$ because when it expands the gas also cools, dropping pressure faster. The ratio is therefore always exactly $\gamma$. ✔️ Answer ::: isotherm $-0.4$, adiabat $-0.56$, ratio $1.4$.

Compression heats the gas, so $T_2>T_1$, making $(T_1-T_2)<0$ and hence $W<0$. $W$ is work done by the gas; $W<0$ means work is done on the gas — exactly what compressing it requires. The energy pushed in raises internal energy, so $T$ rises. Every arrow points the same way. ✔️ Answer ::: $W<0$ (work done on gas), consistent with heating.

Stage A (adiabatic, use $T{V}^{\gamma -1}$): $\gamma -1=2/3$. $T_1=T_0{\left(\frac{V_0}{V_1}\right)}^{\gamma -1}=300(2/4{)}^{2/3}=300(0.5{)}^{0.6667}.$ $(0.5{)}^{0.6667}=e^{-0.6667\ln 2}=e^{-0.4621}=0.6300$, so $T_1=300\times 0.6300=189.0\text{K}$. Pressure after A, from $P{V}^{\gamma }$: $P_1=P_0(V_0/V_1{)}^{\gamma }=8(0.5{)}^{1.6667}=8e^{-1.6667\ln 2}=8e^{-1.1552}=8\times 0.3150=2.520\text{atm}$. Stage B (isothermal, $T$ fixed, use $PV=\text{const}$): $P_2=P_1\frac{V_1}{V_2}=2.520\times \frac{4}{8}=1.260\text{atm}.$ Why switch relations mid-problem: in Stage A no heat flows so $T$ changes ($T{V}^{\gamma -1}$ rules); in Stage B temperature is held, so only $PV=\text{const}$ applies. Using the same relation for both would be wrong physics. ✔️ Answer ::: $T_1\approx 189\text{K}$, $P_2\approx 1.26\text{atm}$.

$v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{1.4\times 1.01\times 10^5}{1.29}}=\sqrt{\frac{1.414\times 10^5}{1.29}}=\sqrt{1.096\times 10^5}.$ $\sqrt{1.096\times 10^5}=331.1\text{m/s}$. Why $\gamma$ and not just $P$: sound compressions are too fast for heat to leak between crests and troughs, so each little compression is adiabatic — pressure responds with the extra factor $\gamma$. Using isothermal $B=P$ (Newton's original error) would predict $\approx 280\text{m/s}$, too slow. ✔️ Answer ::: $v\approx 331\text{m/s}$.

The exponent in $T{V}^{\gamma -1}$ equals $0.4$, so $\gamma -1=0.4\Rightarrow \gamma =1.4$. From $\gamma =1+R/C_V$ (rearranged from $R/C_V=\gamma -1$): $C_V=\frac{R}{\gamma -1}=\frac{R}{0.4}=2.5R$. Then $\frac{f}{2}R=2.5R\Rightarrow f=5$. Reading it: $\gamma =1.4$, $C_V=\frac{5}{2}R$, $f=5$ — a diatomic gas (3 translational + 2 rotational degrees of freedom). The measured curve exponent literally counts the gas's freedoms. ✔️ Answer ::: $\gamma =1.4$, $C_V=2.5R$, $f=5$ (diatomic).

Along the adiabat $P{V}^{\gamma }=P_1{V}_1^{\gamma }\equiv K$, so $P=K{V}^{-\gamma }$. Then $W={\int }_{V_1}^{V_2}K{V}^{-\gamma }dV=K{\left[\frac{V^{1-\gamma }}{1-\gamma }\right]}_{V_1}^{V_2}=\frac{K\left(V_2^{1-\gamma }-V_1^{1-\gamma }\right)}{1-\gamma }.$ Now $K{V}_2^{1-\gamma }=P_2{V}_2^{\gamma }{V}_2^{1-\gamma }=P_2{V}_2$, and likewise $K{V}_1^{1-\gamma }=P_1{V}_1$. So $W=\frac{P_2{V}_2-P_1{V}_1}{1-\gamma }=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}.$ Match to energy method: using $PV=nRT$, $P_1{V}_1-P_2{V}_2=nR(T_1-T_2)$, and $\frac{nR}{\gamma -1}=n{C}_V$ (since $R/C_V=\gamma -1$). Hence $W=n{C}_V(T_1-T_2)$. Two routes, one answer — the integral is the internal-energy drop. ✔️ Numerics: $W=\frac{500-320}{1.4-1}=\frac{180}{0.4}=450\text{J}$. Answer ::: $W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=450\text{J}$.

(a) From $P{V}^{\gamma }=\text{const}$: $\frac{V_2}{V_1}={\left(\frac{P_1}{P_2}\right)}^{1/\gamma }=(2{)}^{1/(5/3)}=2^{0.6}=e^{0.6\ln 2}=e^{0.4159}=1.5157$. (b) From the $TP$ relation, $T_2=T_1{\left(\frac{P_2}{P_1}\right)}^{(\gamma -1)/\gamma }$, with $(\gamma -1)/\gamma =(2/3)/(5/3)=0.4$: $T_2=600(1/2{)}^{0.4}=600e^{-0.4\ln 2}=600e^{-0.2773}=600\times 0.7579=454.7\text{K}.$ (c) No heat flows, so all the work comes from the internal-energy drop: $W=n{C}_V(T_1-T_2)$. With $n=1$ and $C_V=\frac{3}{2}R=1.5\times 8.314=12.471{\text{J K}}^{-1}$, $W=1\times 12.471\times (600-454.7)=12.471\times 145.3=1812\text{J}.$ Consistency chain: pressure halved → volume grew by $\approx 1.52\times$ → temperature fell to $455\text{K}$ → positive work $1812\text{J}$ paid from that cooling. Every relation ($P{V}^{\gamma }$, $TP$, energy) agrees. ✔️ Answer ::: $V_2/V_1\approx 1.516$, $T_2\approx 455\text{K}$, $W\approx 1812\text{J}$.

Adiabatic expansion must cool the gas ($Q=0$, gas spends its own energy doing work), so any reported temperature rise on expansion is impossible. Quantitatively: $T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300(1/2{)}^{0.4}=300e^{-0.4\ln 2}=300\times 0.7579=227.4\text{K}.$ The gas should have cooled to about $227\text{K}$, not warmed to $340\text{K}$. The student likely used the compression exponent sign or swapped $V_1/V_2$. ✔️ Answer ::: impossible; correct $T_2\approx 227\text{K}$.