Homework 6

5-4

(a)

$A\rightarrow B$

The internal energy increases (+) since the value $PV (=nRT)$ increases and the internal energy is only proportional to the temperature. The work is positive (+) since $\int P~dV$ is positive in this process. From $\Delta E_{int}= Q-W>0$, we find $Q>0$ (heat +).

$B\rightarrow C$

The internal energy increases (+) since the value $PV$ increases. The work is zero due to no volume change (0). In this kind of isovolumetric processes, $Q= \Delta E_{int}$. Therefore, $Q>0$ (+).

$C\rightarrow A$

The internal energy decreases (-) for the reason explained above. The work is negative because the volume decreased. From $\Delta E_{int}= Q-W<0$ and $W<0$, we find that $Q<0$.

(b) Firstly, we determine that the sign of the work done by the gas is negative since $\int P~dV <0$ over the process. The value of the work is simply the area enclosed by the process in the $P-V$ diagram: $W = {1\over 2}\times 20 ~\rm Pa \times 2 ~\rm cm^3 = 20 \times 10^{-6} ~\rm J$.

5-5

From $i$ to $f$, the path-independent internal energy increase is $\Delta U_{if} = Q-W = 30 ~\rm cal$. (a) Along $ibf$, $Q = 36~\rm cal$. So $W = Q- \Delta U_{if} = 6 ~\rm cal$. (b) For $fi$, $\Delta U_{fi}=-\Delta U_{if} = -30 ~\rm cal$ and $W= -13 ~\rm cal$. So $Q = \Delta E +W = -43 ~\rm cal$. (c) Since $U_f - U_i = \Delta E = 30 ~\rm cal$, $U_f = 40 ~\rm cal$. (d) For $ibf$, the only work is done over $ib$ since $bf$ is an isovolumetric process, so $W_{ib} = 6 ~\rm cal$. Since we are given that $U_{ib} = 22-10 = 12 ~\rm cal$, $Q_{ib}= 12+6=18 ~\rm cal$. For $bf$, $\Delta U_{bf}=Q_{bf}= 40-22 = 18 ~\rm cal$.

5-6

Let mass of the piston $m_p=2.0 ~\rm kg$, area of the cylinder $A = 2.0 ~\rm cm^2$, atmospheric pressure $P_a = 1.0 ~\rm atm$, density of the steam $\rho = 6.0 \times 10^{-4} ~\rm g/cm^3$, the latent heat of vapourisation $L_v = 2.26 \times 10^6 ~\rm J/kg$, and the rate of contraction $r = 0.3~\rm cm/s$. Then the total pressure on the steam is $P_T = P_a +m_pg/A = 2.0\times 10^5 ~\rm Pa$. (a) Neglecting the volume of water compared to that of steam per unit mass, the volume contraction (condensation) rate is $\alpha = rA$, and the mass condensation rate is $\alpha \rho = 3.6\times 10^{-4} ~\rm g /s$. (b) Heat loss occurs through the loss of the latent heat: $dQ/dt = -\alpha \rho L_v = -8.14\times 10^{-1} ~\rm J/s$. (c) The work rate is $dW/dt = -P_T \alpha = -0.12 ~\rm J/s$. Therefore, the internal energy change rate is $dE/dt = dQ/dt -dW/dt =-0.69 ~\rm J/s$.

5-7

The power transfer per unit area is ${\cal P}/A = 54 ~\rm mW/m^2$, and the thermal conductivity is $k=2.5 ~\rm W/m\cdot K$. Since ${\cal P}/A= k(T_1-T_2)/L$, at $L=35~\rm km$ and $T_2 = 10 ~\rm C$, $T_1 = T_2 +{\cal P}L/(Ak) = 766 ~\rm C$.

5-8

From ${\cal P}/A = k\Delta T/d$, (a) for a glass, $k= 0.8 ~\rm W/m\cdot C$, and the temperature difference is $\Delta T = 92 \times 5/9 = 51 ~\rm C$. So the heat loss rate is $1.36 \times 10^4 ~\rm W$. (b) From the formula ${\cal P}/A = { \Delta T \over \sum_i d_i/k_i}$, for a glass $d/k = 3.75\times 10^{-3}~\rm W^{-1}$ and that of the air gap of $7.5~\rm cm$ is $d/k = 3.2 ~\rm W^{-1}$. So the thermal insulation is mostly done by the air-gap. ${\cal P}/A = 15.9 ~\rm W/m^2$.

6-1

(a) Here the temperature is $293~\rm K$ and $P=1.01\times 10^5 ~\rm Pa$. From $P=\rho k_BT$, where $\rho$ is the number density, $\rho = 2.49\times 10^25 ~\rm m^{-3}$. (b) Molar mass of $\rm N_2$: 28 g. Molar mass of $\rm O_2$: 32 g. If 25 % is oxygen and 75 % is nitrogen, the average molar mass is $m=29 ~\rm g$. So the total mass is $m\rho /N = 1.2\times 10^3 ~\rm g$.

6-2

From the impulse-momenta relationship, $F\Delta t = \Delta p$. When one molecule hits the wall, the momentum transfer to the wall is twice the initial momentum normal to the wall, because the normal component of the momentum of the molecule is reversed in sign. The initial normal momentum of an individual molecule: $p_n = mv_0\cos 55^{\rm o}= 1.89\times 10^{-24} ~\rm kg\cdot m/s$. Since $\Delta p = 2p_n$, and there are $N$ such molecule that hits the area $A$ per second, $P = N\Delta p /(A \Delta t)=1.89 \times 10^3 ~\rm Pa$.

6-3

Average translational kinetic energy: ${1\over 2}m \bar{v^2} ={3\over 2} k_BT= 1.0~\rm eV$. This corresponds to $T=7.7 \times 10^3 ~\rm K$.

6-4

The molar mass: $m = 28.9 ~\rm g$. Molar specific heat: $C_V = 5R/2 = 20.8 ~\rm J/mol\cdot K$. (a) Since one mole is $28.9 ~\rm g$, the constant volume specific heat per unit mass is $7.19\times 10^2 ~\rm J/kg\cdot K$. (b) Mass of the air: From $PV = nRT$, we find that there is 28 mol of air, which corresponds to 0.81 kg. (c) From $\Delta E= MC_V\Delta T$, we need $2.3\times 10^5 ~\rm J$ of energy. (d) If the piston is free to move, the air will move in such a way that the pressure is constant. Since the ration $C_V/C_P = 5/3$, we need energy input of $2.3\times 10^5 \times 3/5= 1.40 \times 10^5~\rm J$.

6-5

Let $P_i,V_i,T_i$ be the initial values of the pressure, volume and temperature, and $P_f, V_f,T_f$ be the final values. (a) From the relation for an adiabatical process $P_iV_i^\gamma =P_fV_f^\gamma$, the final pressure is $P_i (V_i/V_f)^\gamma = 13.6 ~\rm atm$. (b) From $PV = nRT$, we find that $T_f/T_i = {P_fV_f\over P_iV_i}$, so $T_f = 621 ~\rm K$.