HW #2: Cn 19 #'s $6,12,15,30,36,52$$6,12,15,30,36,52$6,12,15,30,36,526,12,15,30,36,52$6,12,15,30,36,52$
19-6) Ist law of TD tells us that the change in total internal energy of the system, $\mathrm{\Delta }U=q+w$$\mathrm{\Delta }U=q+w$Delta U=q+w\Delta U=q+w$\mathrm{\Delta }U=q+w$
or that a small change $dU=\delta q+\delta w$$dU=\delta q+\delta w$dU=delta q+delta wd U=\delta q+\delta w$dU=\delta q+\delta w$ where $q>0$$q>0$q > 0q>0$q>0$ : heat flows into the system
w>0: work done on the system
key to remember: $\cdot W$$\cdot W$*W\cdot W$\cdot W$ is minimiud when the prouss is reversible, ${\omega }_{min}={\omega }_{\text{rev}}$${\omega }_{min}={\omega }_{\text{rev }}$omega_(min)=omega_("rev "){\omega_{m i n}}=\omega_{\text {rev }}${\omega }_{min}={\omega }_{\text{rev }}$

Isotherm: constant $T$$T$TT$T$
$W_{\text{reV}}=-{\int }_{V_1}^{V_2}{P}_{\text{ext}}dV=-{\int }_{V_1}^{V_2}{P}_{\text{int}}dV=-{\int }_{V_1}^{V_2}\frac{nRT}{V}dV=-nRT{\int }_{V_1}^{V_2}\frac{1}{V}dV=-{nRT\ln V|}_{V_1}^{V_2}=-nRT\ln \left(\frac{V_2}{V_1}\right)=nRT\ln \left(\frac{V_1}{V_2}\right)$$W_{\text{reV }}=-{\int }_{V_1}^{V_2} P_{\text{ext }}dV=-{\int }_{V_1}^{V_2} P_{\text{int }}dV=-{\int }_{V_1}^{V_2} \frac{nRT}{V}dV=-nRT{\int }_{V_1}^{V_2} \frac{1}{V}dV=-{\left.nRT\ln V\right|}_{V_1}^{V_2}=-nRT\ln \left(\frac{V_2}{V_1}\right)=nRT\ln \left(\frac{V_1}{V_2}\right)$W_("reV ")=-int_(V_(1))^(V_(2))P_("ext ")dV=-int_(V_(1))^(V_(2))P_("int ")dV=-int_(V_(1))^(V_(2))(nRT)/(V)dV=-nRTint_(V_(1))^(V_(2))(1)/(V)dV=-nRT ln V|_(V_(1))^(V_(2))=-nRT ln((V_(2))/(V_(1)))=nRT ln((V_(1))/(V_(2)))W_{\text {reV }}=-\int_{V_{1}}^{V_{2}} P_{\text {ext }} d V=-\int_{V_{1}}^{V_{2}} P_{\text {int }} d V=-\int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V=-n R T \int_{V_{1}}^{V_{2}} \frac{1}{V} d V=-\left.n R T \ln V\right|_{V_{1}} ^{V_{2}}=-n R T \ln \left(\frac{V_{2}}{V_{1}}\right)=n R T \ln \left(\frac{V_{1}}{V_{2}}\right)$W_{\text{reV }}=-{\int }_{V_1}^{V_2}{P}_{\text{ext }}dV=-{\int }_{V_1}^{V_2}{P}_{\text{int }}dV=-{\int }_{V_1}^{V_2}\frac{nRT}{V}dV=-nRT{\int }_{V_1}^{V_2}\frac{1}{V}dV=-{nRT\ln V|}_{V_1}^{V_2}=-nRT\ln \left(\frac{V_2}{V_1}\right)=nRT\ln \left(\frac{V_1}{V_2}\right)$
$n=5.00$$n=5.00$n=5.00n=5.00$n=5.00$ moles
$T=300\mathrm{k}$$T=300\mathrm{k}$T=300kT=300 \mathrm{k}$T=300\mathrm{k}$
$V_1=100.0{\mathrm{dm}}^3{V}_2=40.0{\mathrm{dm}}^3$$V_1=100.0{\mathrm{dm}}^3{V}_2=40.0{\mathrm{dm}}^3$V_(1)=100.0dm^(3)quadV_(2)=40.0dm^(3)V_{1}=100.0 \mathrm{dm}^{3} \quad V_{2}=40.0 \mathrm{dm}^{3}$V_1=100.0{\mathrm{dm}}^3{V}_2=40.0{\mathrm{dm}}^3$
$w_{\text{min}}=w_{\text{rev}}=-n\mathrm{ATln}\left(\frac{V_1}{V_2}\right)=-5.00\mathrm{cog}\left(\frac{8.314\mathrm{J}}{\sin k}\right)(300\mathrm{K})\underset{=\ln (0.400)}{\underbrace{\ln \left[\frac{40.0\mathrm{dm}{\mathrm{d}}^8}{100.00{\mathrm{dm}}^3}\right]}}=+11,400\mathrm{J}=11.4\mathrm{kJ}$$w_{\text{min }}=w_{\text{rev }}=-n\mathrm{ATln}\left(\frac{V_1}{V_2}\right)=-5.00\mathrm{cog}\left(\frac{8.314\mathrm{J}}{\sin k}\right)(300\mathrm{K})\underset{=\ln (0.400)}{\underbrace{\ln \left[\frac{40.0\mathrm{dm}{\mathrm{d}}^8}{100.00{\mathrm{dm}}^3}\right]}}=+11,400\mathrm{J}=11.4\mathrm{kJ}$w_("min ")=w_("rev ")=-n ATln((V_(1))/(V_(2)))=-5.00 cog((8.314(J))/(sin k))(300K)ubrace(ln[(40.0dmd^(8))/(100.00dm^(3))]ubrace)_(=ln(0.400))=+11,400J=11.4kJw_{\text {min }}=w_{\text {rev }}=-n \operatorname{ATln}\left(\frac{V_{1}}{V_{2}}\right)=-5.00 \operatorname{cog}\left(\frac{8.314 \mathrm{~J}}{\sin k}\right)(300 \mathrm{~K}) \underbrace{\ln \left[\frac{40.0 \mathrm{dm} \mathrm{d}^{8}}{100.00 \mathrm{dm}^{3}}\right]}_{=\ln (0.400)}=+11,400 \mathrm{~J}=11.4 \mathrm{~kJ}$w_{\text{min }}=w_{\text{rev }}=-n\mathrm{ATln}\left(\frac{V_1}{V_2}\right)=-5.00\mathrm{cog}\left(\frac{8.314\mathrm{J}}{\sin k}\right)(300\mathrm{K})\underset{=\ln (0.400)}{\underbrace{\ln \left[\frac{40.0\mathrm{dm}{\mathrm{d}}^8}{100.00{\mathrm{dm}}^3}\right]}}=+11,400\mathrm{J}=11.4\mathrm{kJ}$

Where there this comes from:

Our general desinition of Work is $W=F\cdot \mathrm{\Delta }{h}^{\prime }$$W=F\cdot \mathrm{\Delta }{h}^{\prime }$W=F*Deltah^(')W=F \cdot \Delta h^{\prime}$W=F\cdot \mathrm{\Delta }{h}^{\prime }$, where $W$$W$WW$W$ has units of $N\cdot m=J$$N\cdot m=J$N*m=JN \cdot m=J$N\cdot m=J$, which are units of energy

Explain neg.sign: define $w>0$$w>0$w > 0w>0$w>0$ as work done on gas
Compression: $V_2<V_1-{\int }_{V_1}^{V_2}{P}_{\text{ext}}dV=-P_{\text{ext}}(\underset{(-)}{\underbrace{V_2-V_1}})=-(-)P_{\text{ext}}\mathrm{\Delta }V=P_{\text{ext}}\mathrm{\Delta }V,\uparrow \mathrm{\Delta }U$$V_2<V_1-{\int }_{V_1}^{V_2} P_{\text{ext }}dV=-P_{\text{ext }}(\underset{(-)}{\underbrace{V_2-V_1}})=-(-)P_{\text{ext }}\mathrm{\Delta }V=P_{\text{ext }}\mathrm{\Delta }V,\uparrow \mathrm{\Delta }U$V_(2) < V_(1)-int_(V_(1))^(V_(2))P_("ext ")dV=-P_("ext ")(ubrace(V_(2)-V_(1)ubrace)_((-)))=-(-)P_("ext ")Delta V=P_("ext ")Delta V,quad uarr Delta UV_{2}<V_{1}-\int_{V_{1}}^{V_{2}} P_{\text {ext }} d V=-P_{\text {ext }}(\underbrace{V_{2}-V_{1}}_{(-)})=-(-) P_{\text {ext }} \Delta V=P_{\text {ext }} \Delta V, \quad \uparrow \Delta U$V_2<V_1-{\int }_{V_1}^{V_2}{P}_{\text{ext }}dV=-P_{\text{ext }}(\underset{(-)}{\underbrace{V_2-V_1}})=-(-)P_{\text{ext }}\mathrm{\Delta }V=P_{\text{ext }}\mathrm{\Delta }V,\uparrow \mathrm{\Delta }U$, work done on gas
Expansion: $V_2>V_1-{\int }_{V_1}^{V_2}{p}_{\text{ext}}dV=-p_{\text{ext}}(\underset{(+)}{\underbrace{V_2-V_1}})=-p_{\text{ext}}\mathrm{\Delta }V$$V_2>V_1-{\int }_{V_1}^{V_2} p_{\text{ext }}dV=-p_{\text{ext }}(\underset{(+)}{\underbrace{V_2-V_1}})=-p_{\text{ext }}\mathrm{\Delta }V$V_(2) > V_(1)-int_(V_(1))^(V_(2))p_("ext ")dV=-p_("ext ")(ubrace(V_(2)-V_(1)ubrace)_((+)))=-p_("ext ")Delta VV_{2}>V_{1}-\int_{V_{1}}^{V_{2}} p_{\text {ext }} d V=-p_{\text {ext }}(\underbrace{V_{2}-V_{1}}_{(+)})=-p_{\text {ext }} \Delta V$V_2>V_1-{\int }_{V_1}^{V_2}{p}_{\text{ext }}dV=-p_{\text{ext }}(\underset{(+)}{\underbrace{V_2-V_1}})=-p_{\text{ext }}\mathrm{\Delta }V$, $\downarrow \mathrm{\Delta }U$$\downarrow \mathrm{\Delta }U$darr Delta U\downarrow \Delta U$\downarrow \mathrm{\Delta }U$, gus does work

19-12) $P_1=2.00\mathrm{bar}{P}_2=4.00\mathrm{bar}\frac{P}{V}=C=$$P_1=2.00\mathrm{bar}{P}_2=4.00\mathrm{bar}\frac{P}{V}=C=$quadP_(1)=2.00barquadP_(2)=4.00barquad(P)/(V)=C=\quad P_{1}=2.00 \mathrm{bar} \quad P_{2}=4.00 \mathrm{bar} \quad \frac{P}{V}=C=$P_1=2.00\mathrm{bar}{P}_2=4.00\mathrm{bar}\frac{P}{V}=C=$ constant * $q$$q$qq$q$ and $w$$w$ww$w$ are not state frns. Ther values are path-dependent

Using ideal gas law, can find $V_1,V_2,T_2$$V_1,V_2,T_2$V_(1),V_(2),T_(2)V_{1}, V_{2}, T_{2}$V_1,V_2,T_2$, and $C$$C$CC$C$ : $PV=nRT$$PV=nRT$quad PV=nRT\quad P V=n R T$PV=nRT$

$\frac{P_1}{V_1}=C=\frac{2.00\mathrm{bor}}{0.0113{\mathrm{m}}^2}\cdot \frac{{10}^5\mathrm{kg}}{1\mathrm{bar}\cdot {\mathrm{m}}^2}=1.762\times {10}^7\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}$$\frac{P_1}{V_1}=C=\frac{2.00\mathrm{bor}}{0.0113{\mathrm{m}}^2}\cdot \frac{{10}^5\mathrm{kg}}{1\mathrm{bar}\cdot {\mathrm{m}}^2}=1.762\times {10}^7\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}$(P_(1))/(V_(1))=C=(2.00bor)/(0.0113m^(2))*(10^(5)(kg))/(1bar*m^(2))=1.762 xx10^(7)((kg))/(m^(3)*s^(2))\frac{P_{1}}{V_{1}}=C=\frac{2.00 \mathrm{bor}}{0.0113 \mathrm{~m}^{2}} \cdot \frac{10^{5} \mathrm{~kg}}{1 \mathrm{bar} \cdot \mathrm{m}^{2}}=1.762 \times 10^{7} \frac{\mathrm{~kg}}{\mathrm{~m}^{3} \cdot \mathrm{~s}^{2}}$\frac{P_1}{V_1}=C=\frac{2.00\mathrm{bor}}{0.0113{\mathrm{m}}^2}\cdot \frac{{10}^5\mathrm{kg}}{1\mathrm{bar}\cdot {\mathrm{m}}^2}=1.762\times {10}^7\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}$
$V_2=\frac{P_2}{C}=\frac{4.00\mathrm{bat}-{\mathrm{m}}^2\cdot {\mathrm{s}}^2}{1.762+{10}^7\mathrm{kg}}\cdot \frac{{10}^5\mathrm{kg}}{1.00\mathrm{bar}\cdot \mathrm{gr}\cdot {\mathrm{g}}^2}=0.0226{\mathrm{m}}^2$$V_2=\frac{P_2}{C}=\frac{4.00\mathrm{bat}-{\mathrm{m}}^2\cdot {\mathrm{s}}^2}{1.762+{10}^7\mathrm{kg}}\cdot \frac{{10}^5\mathrm{kg}}{1.00\mathrm{bar}\cdot \mathrm{gr}\cdot {\mathrm{g}}^2}=0.0226{\mathrm{m}}^2$V_(2)=(P_(2))/(C)=(4.00bat-m^(2)*s^(2))/(1.762+10^(7)(kg))*(10^(5)(kg))/(1.00bar*gr*g^(2))=0.0226m^(2)V_{2}=\frac{P_{2}}{C}=\frac{4.00 \mathrm{bat}-\mathrm{m}^{2} \cdot \mathrm{~s}^{2}}{1.762+10^{7} \mathrm{~kg}} \cdot \frac{10^{5} \mathrm{~kg}}{1.00 \mathrm{bar} \cdot \mathrm{gr} \cdot \mathrm{g}^{2}}=0.0226 \mathrm{~m}^{2}$V_2=\frac{P_2}{C}=\frac{4.00\mathrm{bat}-{\mathrm{m}}^2\cdot {\mathrm{s}}^2}{1.762+{10}^7\mathrm{kg}}\cdot \frac{{10}^5\mathrm{kg}}{1.00\mathrm{bar}\cdot \mathrm{gr}\cdot {\mathrm{g}}^2}=0.0226{\mathrm{m}}^2$
$T_2=\frac{P_2{V}_2}{nR}=\frac{4.00\mathrm{bark}\left(0.0226{\mathrm{m}}^2\right)\mathrm{mot}\cdot \mathrm{k}\cdot \frac{{10}^5\mathrm{kg}}{16\mathrm{ar}\cdot {\mathrm{gr}}^2}}{1.00\mathrm{mc}\cdot 8.314\mathrm{z}}\cdot \frac{1\mathrm{J}\cdot {\mathrm{s}}^2}{\mathrm{Kgm}}=1092\mathrm{K}$$T_2=\frac{P_2{V}_2}{nR}=\frac{4.00\mathrm{bark}\left(0.0226{\mathrm{m}}^2\right)\mathrm{mot}\cdot \mathrm{k}\cdot \frac{{10}^5\mathrm{kg}}{16\mathrm{ar}\cdot {\mathrm{gr}}^2}}{1.00\mathrm{mc}\cdot 8.314\mathrm{z}}\cdot \frac{1\mathrm{J}\cdot {\mathrm{s}}^2}{\mathrm{Kgm}}=1092\mathrm{K}$T_(2)=(P_(2)V_(2))/(nR)=(4.00bark(0.0226m^(2))mot*k*(10^(5)(kg))/(16ar*gr^(2)))/(1.00mc*8.314z)*(1(J)*s^(2))/(Kgm)=1092KT_{2}=\frac{P_{2} V_{2}}{n R}=\frac{4.00 \mathrm{bark}\left(0.0226 \mathrm{~m}^{2}\right) \mathrm{mot} \cdot \mathrm{k} \cdot \frac{10^{5} \mathrm{~kg}}{16 \mathrm{ar} \cdot \mathrm{gr}^{2}}}{1.00 \mathrm{mc} \cdot 8.314 \mathrm{z}} \cdot \frac{1 \mathrm{~J} \cdot \mathrm{~s}^{2}}{\mathrm{Kgm}}=1092 \mathrm{~K}$T_2=\frac{P_2{V}_2}{nR}=\frac{4.00\mathrm{bark}\left(0.0226{\mathrm{m}}^2\right)\mathrm{mot}\cdot \mathrm{k}\cdot \frac{{10}^5\mathrm{kg}}{16\mathrm{ar}\cdot {\mathrm{gr}}^2}}{1.00\mathrm{mc}\cdot 8.314\mathrm{z}}\cdot \frac{1\mathrm{J}\cdot {\mathrm{s}}^2}{\mathrm{Kgm}}=1092\mathrm{K}$
$\mathrm{\Delta }U=n{\int }_{T_1}^{T_2}\overline{C_V}dT={n{\overline{C}}_{V}T|}_{T_1}^{T_2}n{\overline{C}}_V(T_2-T_1)=1.00\mathrm{mol}(12.5\mathrm{J}\mathrm{J}/\mathrm{K})(1092\mathrm{K}-273\mathrm{K})=10,200\mathrm{J}=10.2\mathrm{kJ}(C_V=n\overline{C_V})$$\mathrm{\Delta }U=n{\int }_{T_1}^{T_2} \overline{C_V}dT={\left.n{\overline{C}}_{V}T\right|}_{T_1}^{T_2}n{\overline{C}}_V\left(T_2-T_1\right)=1.00\mathrm{mol}(12.5\mathrm{J}\mathrm{J}/\mathrm{K})(1092\mathrm{K}-273\mathrm{K})=10,200\mathrm{J}=10.2\mathrm{kJ}\left(C_V=n\overline{C_V}\right)$Delta U=nint_(T_(1))^(T_(2)) bar(C_(V))dT=n bar(C)_(V)T|_(T_(1))^(T_(2))n bar(C)_(V)(T_(2)-T_(1))=1.00mol(12.5JJ//K)(1092K-273K)=10,200J=10.2kJquad(C_(V)=n bar(C_(V)))\Delta U=n \int_{T_{1}}^{T_{2}} \overline{C_{V}} d T=\left.n \bar{C}_{V} T\right|_{T_{1}} ^{T_{2}} n \bar{C}_{V}\left(T_{2}-T_{1}\right)=1.00 \mathrm{~mol}(12.5 \mathrm{~J} \mathrm{~J} / \mathrm{K})(1092 \mathrm{~K}-273 \mathrm{~K})=10,200 \mathrm{~J}=10.2 \mathrm{~kJ} \quad\left(C_{V}=n \overline{C_{V}}\right)$\mathrm{\Delta }U=n{\int }_{T_1}^{T_2}\overline{C_V}dT={n{\overline{C}}_{V}T|}_{T_1}^{T_2}n{\overline{C}}_V(T_2-T_1)=1.00\mathrm{mol}(12.5\mathrm{J}\mathrm{J}/\mathrm{K})(1092\mathrm{K}-273\mathrm{K})=10,200\mathrm{J}=10.2\mathrm{kJ}(C_V=n\overline{C_V})$
$w=-{\int }_{V_1}^{V_2}{p}_{\text{ext}}dV=-{\int }_{V_1}^{V_2}{p}_{\text{int}}dV=-\sum _{V_1}^{V_2}\underset{\text{sub}{p}_{\text{int}}=CV}{\underbrace{CV}}dV=-{\frac{C{V}^2}{2}|}_{V_1}^{V_2}=\frac{-C}{2}(V_2^2-V_1^2)=\frac{-1.762\times {10}^2\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}}{2}({\left(0.0226{\mathrm{m}}^2\right)}^2-{\left(0.0113{\mathrm{m}}^2\right)}^2)=$$w=-{\int }_{V_1}^{V_2} p_{\text{ext }}dV=-{\int }_{V_1}^{V_2} p_{\text{int }}dV=-\sum _{V_1}^{V_2} \underset{\text{sub }{p}_{\text{int }}=CV}{\underbrace{CV}}dV=-{\left.\frac{C{V}^2}{2}\right|}_{V_1}^{V_2}=\frac{-C}{2}\left(V_2^2-V_1^2\right)=\frac{-1.762\times {10}^2\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}}{2}\left({\left(0.0226{\mathrm{m}}^2\right)}^2-{\left(0.0113{\mathrm{m}}^2\right)}^2\right)=$w=-int_(V_(1))^(V_(2))p_("ext ")dV=-int_(V_(1))^(V_(2))p_("int ")dV=-sum_(V_(1))^(V_(2))ubrace(CVubrace)_("sub "p_("int ")=CV)dV=-(CV^(2))/(2)|_(V_(1))^(V_(2))=(-C)/(2)(V_(2)^(2)-V_(1)^(2))=(-1.762 xx10^(2)((kg))/(m^(3)*s^(2)))/(2)((0.0226m^(2))^(2)-(0.0113m^(2))^(2))=w=-\int_{V_{1}}^{V_{2}} p_{\text {ext }} d V=-\int_{V_{1}}^{V_{2}} p_{\text {int }} d V=-\sum_{V_{1}}^{V_{2}} \underbrace{C V}_{\text {sub } p_{\text {int }}=C V} d V=-\left.\frac{C V^{2}}{2}\right|_{V_{1}} ^{V_{2}}=\frac{-C}{2}\left(V_{2}^{2}-V_{1}^{2}\right)=\frac{-1.762 \times 10^{2} \frac{\mathrm{~kg}}{\mathrm{~m}^{3} \cdot \mathrm{~s}^{2}}}{2}\left(\left(0.0226 \mathrm{~m}^{2}\right)^{2}-\left(0.0113 \mathrm{~m}^{2}\right)^{2}\right)=$w=-{\int }_{V_1}^{V_2}{p}_{\text{ext }}dV=-{\int }_{V_1}^{V_2}{p}_{\text{int }}dV=-\sum _{V_1}^{V_2}\underset{\text{sub }{p}_{\text{int }}=CV}{\underbrace{CV}}dV=-{\frac{C{V}^2}{2}|}_{V_1}^{V_2}=\frac{-C}{2}(V_2^2-V_1^2)=\frac{-1.762\times {10}^2\frac{\mathrm{kg}}{{\mathrm{m}}^3\cdot {\mathrm{s}}^2}}{2}({\left(0.0226{\mathrm{m}}^2\right)}^2-{\left(0.0113{\mathrm{m}}^2\right)}^2)=$

$\mathrm{\Delta }U=q+w\Rightarrow q=\mathrm{\Delta }U-w=10.2\mathrm{kJ}-(-3.40\mathrm{kJ})=13.6\mathrm{kJ}$$\mathrm{\Delta }U=q+w\Rightarrow q=\mathrm{\Delta }U-w=10.2\mathrm{kJ}-(-3.40\mathrm{kJ})=13.6\mathrm{kJ}$Delta U=q+w=>q=Delta U-w=10.2kJ-(-3.40kJ)=13.6kJ\Delta U=q+w \Rightarrow q=\Delta U-w=10.2 \mathrm{~kJ}-(-3.40 \mathrm{~kJ})=13.6 \mathrm{~kJ}$\mathrm{\Delta }U=q+w\Rightarrow q=\mathrm{\Delta }U-w=10.2\mathrm{kJ}-(-3.40\mathrm{kJ})=13.6\mathrm{kJ}$
Enthalpy is a quantity we define as: $H=U+\underset{se+PV}{\underbrace{}}=$$H=U+\underset{se+PV}{\underbrace{}}=$H=U+ubrace(ubrace)_(se+PV)=H=U+\underbrace{}_{s e+P V}=$H=U+\underset{se+PV}{\underbrace{}}=$ RAT
$\mathrm{\Delta }H=\mathrm{\Delta }U+P\mathrm{\Delta }V+V\mathrm{\Delta }P=\underset{T_{\text{easher to compute}}}{\underbrace{\mathrm{\Delta }U+\text{nRAT}}}$$\mathrm{\Delta }H=\mathrm{\Delta }U+P\mathrm{\Delta }V+V\mathrm{\Delta }P=\underset{T_{\text{easher to compute }}}{\underbrace{\mathrm{\Delta }U+\text{ nRAT }}}$Delta H=Delta U+P Delta V+V Delta P=ubrace(Delta U+" nRAT "ubrace)_(T_("easher to compute "))\Delta H=\Delta U+P \Delta V+V \Delta P=\underbrace{\Delta U+\text { nRAT }}_{T_{\text {easher to compute }}}$\mathrm{\Delta }H=\mathrm{\Delta }U+P\mathrm{\Delta }V+V\mathrm{\Delta }P=\underset{T_{\text{easher to compute }}}{\underbrace{\mathrm{\Delta }U+\text{ nRAT }}}$
so $\mathrm{\Delta }H=\mathrm{\Delta }U+$$\mathrm{\Delta }H=\mathrm{\Delta }U+$Delta H=Delta U+\Delta H=\Delta U+$\mathrm{\Delta }H=\mathrm{\Delta }U+$ nR $\mathrm{\Delta }T=10.2\mathrm{kJ}+\underset{6,809.23=6.8\mathrm{kJ}}{\underbrace{1.00\mathrm{mot}\left(8.31\frac{4\mathrm{J}}{K\cdot 908}\right)(1092\mathrm{K}-273\mathrm{K})}}=(10.2+6.8)\mathrm{kJ}=17.0\mathrm{kJ}$$\mathrm{\Delta }T=10.2\mathrm{kJ}+\underset{6,809.23=6.8\mathrm{kJ}}{\underbrace{1.00\mathrm{mot}\left(8.31\frac{4\mathrm{J}}{K\cdot 908}\right)(1092\mathrm{K}-273\mathrm{K})}}=(10.2+6.8)\mathrm{kJ}=17.0\mathrm{kJ}$Delta T=10.2kJ+ubrace(1.00mot(8.31(4(J))/(K*908))(1092(K)-273(K))ubrace)_(6,809.23=6.8kJ)=(10.2+6.8)kJ=17.0kJ\Delta T=10.2 \mathrm{~kJ}+\underbrace{1.00 \mathrm{mot}\left(8.31 \frac{4 \mathrm{~J}}{K \cdot 908}\right)(1092 \mathrm{~K}-273 \mathrm{~K})}_{6,809.23=6.8 \mathrm{~kJ}}=(10.2+6.8) \mathrm{kJ}=17.0 \mathrm{~kJ}$\mathrm{\Delta }T=10.2\mathrm{kJ}+\underset{6,809.23=6.8\mathrm{kJ}}{\underbrace{1.00\mathrm{mot}\left(8.31\frac{4\mathrm{J}}{K\cdot 908}\right)(1092\mathrm{K}-273\mathrm{K})}}=(10.2+6.8)\mathrm{kJ}=17.0\mathrm{kJ}$

19-15) Show $\frac{T_2}{T_1}={\left(\frac{V}{V_2}\right)}^{\frac{\mathrm{\Lambda }}{\overline{C_v}}}$$\frac{T_2}{T_1}={\left(\frac{V}{V_2}\right)}^{\frac{\mathrm{\Lambda }}{\overline{C_v}}}$(T_(2))/(T_(1))=((V)/(V_(2)))^((Lambda)/( bar(C_(v))))\frac{T_{2}}{T_{1}}=\left(\frac{V}{V_{2}}\right)^{\frac{\Lambda}{\overline{C_{v}}}}$\frac{T_2}{T_1}={\left(\frac{V}{V_2}\right)}^{\frac{\mathrm{\Lambda }}{\overline{C_v}}}$ for reversible adiabatic expansion of ideal gas
adiabatic: ${\delta }_q=0$${\delta }_q=0$delta_(q)=0\delta_{q}=0${\delta }_q=0$
$dU=\delta {\xi }^0+\delta w=\delta w\Rightarrow dU=\delta w$$dU=\delta {\xi }^0+\delta w=\delta w\Rightarrow dU=\delta w$dU=deltaxi^(0)+delta w=delta w quad=>dU=delta wd U=\delta \xi^{0}+\delta w=\delta w \quad \Rightarrow d U=\delta w$dU=\delta {\xi }^0+\delta w=\delta w\Rightarrow dU=\delta w$
*want to rewrite $dU$$dU$dUd U$dU$ and $\delta w$$\delta w$delta w\delta w$\delta w$, then work out a relationship:

We can write U (or any state function) as a total disserential:
${U(T,V)\Rightarrow dU=\frac{\partial U}{\partial T})}_{V}dT+{\frac{\partial U}{\partial V}|}_{T}dV$${\left.U(T,V)\Rightarrow dU=\frac{\partial U}{\partial T}\right)}_{V}dT+{\left.\frac{\partial U}{\partial V}\right|}_{T}dV$U(T,V)=>dU=(del U)/(del T))_(V)dT+(del U)/(del V)|_(T)dV quad\left.U(T, V) \Rightarrow d U=\frac{\partial U}{\partial T}\right)_{V} d T+\left.\frac{\partial U}{\partial V}\right|_{T} d V \quad${U(T,V)\Rightarrow dU=\frac{\partial U}{\partial T})}_{V}dT+{\frac{\partial U}{\partial V}|}_{T}dV$ Use: $dU=n{\overline{C}}_{V}dT$$dU=n{\overline{C}}_{V}dT$dU=n bar(C)_(V)dT quadd U=n \bar{C}_{V} d T \quad$dU=n{\overline{C}}_{V}dT$ Q $w_{\text{reV}}=-{\int }_{P_{\text{ext}}}dV=-{\int }_{\text{int}}dV=-\int \frac{nRT}{V}dV$$w_{\text{reV }}=-{\int }_{P_{\text{ext }}} dV=-{\int }_{\text{int }} dV=-\int \frac{nRT}{V}dV$quadw_("reV ")=-int_(P_("ext "))dV=-int_("int ")dV=-int(nRT)/(V)dV\quad w_{\text {reV }}=-\int_{P_{\text {ext }}} d V=-\int_{\text {int }} d V=-\int \frac{n R T}{V} d V$w_{\text{reV }}=-{\int }_{P_{\text{ext }}}dV=-{\int }_{\text{int }}dV=-\int \frac{nRT}{V}dV$
${\frac{\partial U}{\partial N})}_T=0$${\left.\frac{\partial U}{\partial N}\right)}_T=0$(del U)/(del N))_(T)=0\left.\frac{\partial U}{\partial N}\right)_{T}=0${\frac{\partial U}{\partial N})}_T=0$ for ideal gas $\to$$\to$rarr\rightarrow$\to$ asour Uonly changs $w/T:\overline{U}=\frac{3}{2}RT$$w/T:\overline{U}=\frac{3}{2}RT$w//T: bar(U)=(3)/(2)RTw / T: \bar{U}=\frac{3}{2} R T$w/T:\overline{U}=\frac{3}{2}RT$
(each D.O.F contriburs $\frac{RT}{2}$$\frac{RT}{2}$(RT)/(2)\frac{R T}{2}$\frac{RT}{2}$ to $\overline{U}$$\overline{U}$bar(U)\bar{U}$\overline{U}$ )
$\begin{array}{ll}\Rightarrow dU=\delta {\omega }_{\text{reV}}& P_{\text{ext}}=P_{\text{int for}}\text{revesible work}\\ n{\overline{C}}_{V}dT=-P_{\text{int}}d\overline{V}& \text{sub}P=\frac{nRT}{V}\end{array}$$\begin{array}{l}\Rightarrow dU=\delta {\omega }_{\text{reV }}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{P}_{\text{ext }}=P_{\text{int for }}\text{ revesible work }\\ n{\overline{C}}_{V}dT=-P_{\text{int }}d\overline{V}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ sub }P=\frac{nRT}{V}\end{array}${:[=>dU=deltaomega_("reV "),quadP_("ext ")=P_("int for ")" revesible work "],[n bar(C)_(V)dT=-P_("int ")d bar(V)," sub "P=(nRT)/(V)]:}\begin{array}{ll}\Rightarrow d U=\delta \omega_{\text {reV }} & \quad P_{\text {ext }}=P_{\text {int for }} \text { revesible work } \\ n \bar{C}_{V} d T=-P_{\text {int }} d \bar{V} & \text { sub } P=\frac{n R T}{V}\end{array}$\begin{array}{ll}\Rightarrow dU=\delta {\omega }_{\text{reV }}& P_{\text{ext }}=P_{\text{int for }}\text{ revesible work }\\ n{\overline{C}}_{V}dT=-P_{\text{int }}d\overline{V}& \text{ sub }P=\frac{nRT}{V}\end{array}$
$A\overline{C}dT=\frac{-QRT}{V}dV$$A\overline{C}dT=\frac{-QRT}{V}dV$A bar(C)dT=(-QRT)/(V)dVA \bar{C} d T=\frac{-Q R T}{V} d V$A\overline{C}dT=\frac{-QRT}{V}dV$ reacrange
${\int }_{T_1}^{T_2}\frac{\overline{C_V}dT}{T}={\int }_{V_1}^{V_2}\frac{-R}{V}dV$${\int }_{T_1}^{T_2} \frac{\overline{C_V}dT}{T}={\int }_{V_1}^{V_2} \frac{-R}{V}dV$int_(T_(1))^(T_(2))( bar(C_(V))dT)/(T)=int_(V_(1))^(V_(2))(-R)/(V)dV\int_{T_{1}}^{T_{2}} \frac{\overline{C_{V}} d T}{T}=\int_{V_{1}}^{V_{2}} \frac{-R}{V} d V${\int }_{T_1}^{T_2}\frac{\overline{C_V}dT}{T}={\int }_{V_1}^{V_2}\frac{-R}{V}dV$ Integrate B.S.
${\overline{C_V}\ln T|}_{T_1}^{T_2}=-{R\ln V|}_{V_1}^{V_2}$${\left.\overline{C_V}\ln T\right|}_{T_1}^{T_2}=-{\left.R\ln V\right|}_{V_1}^{V_2}$bar(C_(V))ln T|_(T_(1))^(T_(2))=-R ln V|_(V_(1))^(V_(2))\left.\overline{C_{V}} \ln T\right|_{T_{1}} ^{T_{2}}=-\left.R \ln V\right|_{V_{1}} ^{V_{2}}${\overline{C_V}\ln T|}_{T_1}^{T_2}=-{R\ln V|}_{V_1}^{V_2}$
$\overline{C_v\ln \left(\frac{T_2}{T_1}\right)}=-\widehat{R\ln \left(\frac{V_2}{V_1}\right)}$$\overline{C_v\ln \left(\frac{T_2}{T_1}\right)}=-\widehat{R\ln \left(\frac{V_2}{V_1}\right)}$bar(C_(v)ln((T_(2))/(T_(1))))=- widehat(R ln((V_(2))/(V_(1))))\overline{C_{v} \ln \left(\frac{T_{2}}{T_{1}}\right)}=-\widehat{R \ln \left(\frac{V_{2}}{V_{1}}\right)}$\overline{C_v\ln \left(\frac{T_2}{T_1}\right)}=-\widehat{R\ln \left(\frac{V_2}{V_1}\right)}$
$e^{\ln {\left(\frac{T_2}{T_1}\right)}^{\overline{V_v}}=e^{\ln {\left(\frac{V_2}{V_1}\right)}^{-R}}\text{exponentiate B.S.}}$$e^{\ln {\left(\frac{T_2}{T_1}\right)}^{\overline{V_v}}=e^{\ln {\left(\frac{V_2}{V_1}\right)}^{-R}}\text{ exponentiate B.S. }}$e^(ln ((T_(2))/(T_(1)))^( bar(V_(v)))=e^(ln ((V_(2))/(V_(1)))^(-R))quad" exponentiate B.S. ")e^{\ln \left(\frac{T_{2}}{T_{1}}\right)^{\overline{V_{v}}}=e^{\ln \left(\frac{V_{2}}{V_{1}}\right)^{-R}} \quad \text { exponentiate B.S. }}$e^{\ln {\left(\frac{T_2}{T_1}\right)}^{\overline{V_v}}=e^{\ln {\left(\frac{V_2}{V_1}\right)}^{-R}}\text{ exponentiate B.S. }}$

19-30) Fur ideal gas: ${\left(\frac{\partial U}{\partial V}\right)}_T=0$${\left(\frac{\partial U}{\partial V}\right)}_T=0$((del U)/(del V))_(T)=0\left(\frac{\partial U}{\partial V}\right)_{T}=0${\left(\frac{\partial U}{\partial V}\right)}_T=0$, prove ${\left(\frac{\partial H}{\partial V}\right)}_T=0$${\left(\frac{\partial H}{\partial V}\right)}_T=0$((del H)/(del V))_(T)=0\left(\frac{\partial H}{\partial V}\right)_{T}=0${\left(\frac{\partial H}{\partial V}\right)}_T=0$
Recall $H=U+PV\stackrel{\text{ideal gas law}}{=}U+nRT$$H=U+PV\stackrel{\text{ ideal gas law }}{=}U+nRT$H=U+PV=^(" ideal gas law ")U+nRTH=U+P V \stackrel{\text { ideal gas law }}{=} U+n R T$H=U+PV\stackrel{\text{ ideal gas law }}{=}U+nRT$
If keep $H=U+PV$$H=U+PV$H=U+PVH=U+P V$H=U+PV$, need to use product rule on ${(\frac{\partial }{\partial V}(PV))}_T$${\left(\frac{\partial }{\partial V}(PV)\right)}_T$((del)/(del V)(PV))_(T)\left(\frac{\partial}{\partial V}(P V)\right)_{T}${(\frac{\partial }{\partial V}(PV))}_T$ :