3.0 First law of thermodynamics

(a) For this transition, what are the systems change in internal energy?

(b) If the system moves from $B$ to $A$, what is the change in internal energy?

(c) If in moving from $A$ to $B$ along a different path in which $W_{AB}^/ = 400J$ of work is done on the system, how much heat does it absorb?

(a) From the first law,$$\begin{equation} \begin{aligned} \Delta {U_{AB}} = {Q_{AB}} - {W_{AB}} = \left( {400 - 100} \right)J \\ = 300J \\\end{aligned} \end{equation} $$

(b) Consider a closed path that passes through the state $A$ and $B$. Internal energy is a state function so $\Delta U$ is zero for a closed path.

Thus,$$\begin{equation} \begin{aligned} \Delta U = \Delta {U_{AB}} + \Delta {U_{BA}} = 0 \\ \Delta {U_{BA}} = - \Delta {U_{AB}} \\ = - 300J \\\end{aligned} \end{equation} $$

(C) The change in internal energy is the same for any path, so$$\begin{equation} \begin{aligned} \Delta {U_{AB}} = \Delta U_{AB}^/ = Q_{AB}^/ - W_{AB}^/ \\ 300J = Q_{AB}^/ - \left( { - 400J} \right) \\\end{aligned} \end{equation} $$and the heat exchanged is $$Q_{AB}^/ = - 100J$$The negative sign indicates that the system loses heat in this transition.

Question 5. A gas undergoes a process such that $P \propto \frac{1}{T}$. If the molar heat capacity for this process is $C = 33.24\,J/mol - K$. Find the degree of freedom of molecules of the gas.

Solution: As $$\begin{equation} \begin{aligned} P \propto \frac{1}{T} \\ or\quad PT = constant ....(1)\\\end{aligned} \end{equation} $$We have for one mole of an ideal gas $$PV = RT ....(2)$$From the above two equations$$\begin{equation} \begin{aligned} {P^2}V = cons\operatorname{tant} \\ P{V^{1/2}} = K\quad (say) ......(3) \\\end{aligned} \end{equation} $$From first law of thermodynamics,$$\begin{equation} \begin{aligned} \Delta Q = \Delta U + \Delta W \\ C\Delta T = {C_V}\Delta T + \Delta W \\ C = {C_V} + \frac{{\Delta W}}{{\Delta T}} .....(4) \\\end{aligned} \end{equation} $$Here,$$\begin{equation} \begin{aligned} \Delta W = \int {PdV} = K\int\limits_{{V_i}}^{{V_f}} {{V^{ - 1/2}}dV} \\ = \frac{{{P_f}{V_f} - {P_i}{V_i}}}{{1 - \left( {1/2} \right)}} = \frac{{R\left( {{T_f} - {T_i}} \right)}}{{1/2}} = \frac{{R\Delta T}}{{1/2}} \\ \frac{{\Delta W}}{{\Delta T}} = 2R \\\end{aligned} \end{equation} $$Substituting in equation (4) ,We have$$C = {C_V} + 2R = \frac{R}{{\gamma - 1}} + 2R$$Substituting the values,$$\begin{equation} \begin{aligned} 33.24 = R\left( {\frac{1}{{\gamma - 1}} + 2} \right) = 8.31\left( {\frac{1}{{\gamma - 1}} + 2} \right) \\ \gamma = 1.5 \\ As\quad \gamma = 1 + \frac{2}{F} \\ F = \frac{2}{{\gamma - 1}} = \frac{2}{{1.5 - 1}} = 4 \\\end{aligned} \end{equation} $$

Question 6. A gaseous mixture enclosed in a vessel consists of one g mole of a gas A with $\gamma = \left( {\frac{5}{3}} \right)$ and some amount of gas B with $\gamma = \left( {\frac{7}{5}} \right)$ at a temperature $T$.The gases Aand B do not react with each other and are assumed to be ideal. Find the number of g moles of the gas B if $\gamma $ for the gaseous mixture is $\left( {\frac{{19}}{{13}}} \right)$.

Solution: As for an ideal gas, $$\begin{equation} \begin{aligned} {C_P} - {C_V} = R\quad and\;\gamma = \left( {\frac{{{C_P}}}{{{C_V}}}} \right) \\ so,\quad {C_V} = \frac{R}{{\gamma - 1}} \\ \therefore \;{\left( {{C_V}} \right)_1} = \frac{R}{{\left( {\frac{5}{3}} \right) - 1}} = \frac{3}{2}R;\quad {\left( {{C_V}} \right)_2} = \frac{R}{{\left( {\frac{7}{5} - 1} \right)}} = \frac{5}{2}R \\ and\quad {\left( {{C_V}} \right)_{\max }} = \frac{R}{{\left( {\frac{{19}}{{13}}} \right) - 1}} = \frac{{13}}{6}R \\\end{aligned} \end{equation} $$Now from conservation of energy,$$\begin{equation} \begin{aligned} \Delta U = \Delta {U_1} + \Delta {U_2} \\ \left( {{n_1} + {n_2}} \right){\left( {{C_V}} \right)_{mix}}\Delta T = \left[ {{n_1}{{\left( {{C_V}} \right)}_1} + {n_2}{{\left( {{C_V}} \right)}_2}} \right]\Delta T \\ {\left( {{C_V}} \right)_{mix}} = \frac{{\left[ {{n_1}{{\left( {{C_V}} \right)}_1} + {n_2}{{\left( {{C_V}} \right)}_2}} \right]}}{{{n_1} + {n_2}}} \\ \frac{{13}}{6}R = \frac{{1 \times \frac{3}{2}R + {n_2}\frac{5}{2}R}}{{1 + {n_2}}} = \frac{{\left( {3 + 5{n_2}} \right)R}}{{2\left( {1 + {n_2}} \right)}} \\ 13 + 13{n_2} = 9 + 15{n_2}, \\ {n_2} = 2\,g\,mole \\\end{aligned} \end{equation} $$

Question 7. A certain amount of ideal gas passes from state $A$ to $B$ first by means of process $1$, then by means of process $2$. In which of the process is the amount of heat absorbed by the gas greater?

$$\begin{equation} \begin{aligned} {Q_1} = {W_1} + \Delta {U_1} \\ and\;{Q_2} = {W_2} + \Delta {U_2} \\\end{aligned} \end{equation} $$ $U$ is a state function. Hence, $\Delta U$ depends only on the initial and final positions. Therefore,

$$\Delta {U_1} = \Delta {U_2}$$But$${W_1} > {W_2}$$ As the area under 1 is greater than area under 2. Hence,$${Q_1} > {Q_2}$$

Question 8. Calculate the increase in internal energy of $1 kg$ of water at ${100^o}C$ when it is converted into steam at the same temperature and at $1 atm$ ($10 kPa$). The density of the water and steam are $1000\,kg\,{m^{ - 3}}$ and $0.6\,kg\,{m^{ - 3}}$ respectively. The latent heat of vaporisation of water = $2.25\, \times \,{10^6}J\,k{g^{ - 1}}$.

Solution: The volume of $1 kg$ of water= $\frac{1}{{1000}}{m^3}$ and of $1 kg$ of steam=$\frac{1}{{0.6}}{m^3}$.

The increase in volume $$\begin{equation} \begin{aligned} = \frac{1}{{0.6}}{m^3} - \frac{1}{{1000}}{m^3} \\ = \left( {1.7 - 0.001} \right){m^3} \approx \~1.7\,{m^3} \\\end{aligned} \end{equation} $$The work done by the piston is $$\begin{equation} \begin{aligned} = P\Delta V \\ = \left( {100kPa} \right)\left( {1.7{m^3}} \right) \\ = 1.7 \times {10^5}\,J \\\end{aligned} \end{equation} $$The heat given to convert 1 kg of water into steam$$ = 2.25 \times {10^6}\,J$$The change in internal energy is$$\begin{equation} \begin{aligned} \Delta U = \Delta Q - \Delta W \\ = 2.25 \times {10^6}J - 1.7 \times {10^5}J \\ = 2.08 \times {10^6}J \\\end{aligned} \end{equation} $$

Question 9. The internal energy of a monoatomic ideal gas is $1.5$ $nRT$. One mole of helium is kept in a cylinder of crosssection $8.5\,c{m^2}$. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of $42 J$ heat is given to the gas. If the temperature rises through ${2^o}C$, Find the distance moved by the piston. Atmospheric pressure=$100 kPa$.

Solution: The change in internal energy of the gas is $$\begin{equation} \begin{aligned} \Delta U = 1.5\,nR\Delta T \\ = 1.5\left( {1\,mol} \right)\left( {8.3\,J\,{K^{ - 1}}\,mo{l^{ - 1}}} \right)\left( {2\,K} \right) \\ = 24.9J. \\\end{aligned} \end{equation} $$The heat given to the gas =42 J.

The work done by the gas is $$\begin{equation} \begin{aligned} \Delta W = \Delta Q - \Delta U \\ = 42\,J - 24.9\,J = 17.1\,J \\\end{aligned} \end{equation} $$If the distance moved by the piston is x, the work done is $$\Delta W = \left( {100kPa} \right)\left( {8.5\,c{m^2}} \right)x$$Thus$$\begin{equation} \begin{aligned} \left( {{{10}^5}\,N\,{m^{ - 2}}} \right)\left( {8.5 \times {{10}^{ - 4}}{m^{ - 2}}} \right)x = 17.1J \\ x = 0.2m = 20cm \\\end{aligned} \end{equation} $$

Question 10. Suppose $1.0 g$ of water vaporises isobarically at atmospheric pressure $\left( {1.01 \times {{10}^5}Pa} \right)$. Its volume in the liquid state is ${V_i} = {V_{liquid}} = 1.0\,c{m^3}$ and its volume in vapour state is ${V_f} = {V_{vapour}} = 1671\,c{m^3}$. Find the work done in the expansion and the change in internal energy in the system. Ignore any mixing of the steam and the surrounding air. Take latent heat of vaporisation ${L_v} = 2.26 \times {10^6}J/kg$.

Solution: Because the expansion takes place at constant pressure, the work done is $$\begin{equation} \begin{aligned} W = \int\limits_{{V_i}}^{{V_f}} {{P_0}dV} = {p_0}\int\limits_{{V_i}}^{{V_f}} {dV} = {p_0}\left( {{V_f} - {V_i}} \right) \\ = \left( {1.01 \times {{10}^5}} \right)\left( {1671 \times {{10}^{ - 6}} - 1.0 \times {{10}^{ - 6}}} \right) \\ = 169J \\ Q = m{L_v} = \left( {1.0 \times {{10}^{ - 3}}} \right)\left( {2.26 \times {{10}^6}} \right) = 2260J \\\end{aligned} \end{equation} $$Hence, from the first law, the change in internal energy$$\Delta U = Q - W = 2260 - 169 = 2091J$$