HW 7

6-6

(a) First of all, the internal energy change is easy to calculate since it only depends on the temperature change. ( $\Delta U = {3\over 2}nR \Delta T$.) For $1\rightarrow 2$ process, since the volume does not change, the work is zero, and so $\Delta U_{12} = Q_{12} = {3\over 2}nR ( T_2-T_1)=3.74\times 10^3~\rm J$. For $2\rightarrow 3$, there is no exchange of heat (adiabatic) $Q_{23} =0$ which means that the work $W_{23}=-\Delta U_{12} = -{3\over 2}nR ( T_3-T_2)=1.80\times 10^3 ~\rm J$. For $3\rightarrow 1$, $\Delta U_{31}={3\over 2}nR ( T_1-T_3)=-1.93\times 10^3 ~\rm J$. From the differential relation between the volume variation and the internal energy for a constant-pressure process, $PdV = nRdT= {2\over 3}dU$, $W_{31}={2\over 3}\Delta U_{31}=-1.29\times 10^3~\rm J$. Finally, $Q_{31}= W_{31}+\Delta U_{31}=-3.22\times 10^3 ~\rm J$. In the overall process, if you add all the numbers above, $\Delta U=0$, $Q = 5.2\times 10^2 ~\rm J$, and $W=Q= 5.2 \times 10^2 ~\rm J$.

(b) The initial volume is $V_1 =nR/T_1P_1 = 2.74\times 10^{-7}~\rm m^3$. At point 2, since the volume is the same as at point 1, $P_2 = P1 (T_2/T_1) = 2.0 ~\rm atm$. At point 3, since the pressure is the same as at point 1, $V_3 =V_1 (T_3/T_1)=4.15\times 10^{-7}~\rm m^3$.

6-7

From ${1\over 2} m\bar{v^2}={3\over 2}k_BT$ and the mass of the oxygen molecule $32\times 1.67\times 10^{-27} = 5.34\times 10^{-26}~\rm kg$, (a) $T=m\bar{v^2} /3k_B = 1.62\times 10^5 ~\rm K$, and (b) $6.82\times 10^3 ~\rm K$. (c) In case of hydrogen, the escape temperatures is smaller by a factor of 1/16, which gives $1.01\times 10^4 ~\rm K$ for the earth. Since both of the escape temperature is higher than the actual temperature, you would expect to find much oxygen and hydrogen.

6-8

The pressure is $P=1.01\times 10^3 ~\rm Pa$. From $P=\rho_Nk_BT$ where $\rho_N$ is the number density and $\rho_m= m\rho_N$ where $\rho_m$ is the mass density and $m$ is the individual molecular mass, you get $m={\rho_m k_BT\over P}=4.63\times 10^{-26} ~\rm kg$. Then $v_{\rm rms} = \sqrt{3k_BT/m}=4.94\times 10^2 ~\rm m/s$.

7-1

The efficiency is $e=W/Q_h=(Q_h-Q_c)/Q_h =0.250$. (a) The heat expelled is $Q_c = 8000~\rm J$, so $Q_h = Q_c/(1-e) =107\times 10^2~\rm J$. (b) Since the work is $W= Q_h - Q_c = 2.7\times 10^3~\rm J$, the time is $W/{\cal P}=0.540 ~\rm s$.

7-2

(a) The total work done is simply the area enclosed by the process in the $P-V$ diagram. $W=\oint P~ dV = (P-P_0)(V-V_0)=2.27\times 10^ 3 ~\rm J$. (b) Firstly, the internal energy difference between $c$ and $a$ is $\Delta U _{ac}={3\over 2}( PV-P_0V_0)=1.02\times 10^4 ~\rm J$. The work done (over $bc$ only) is $W_{ac}=W_{bc}=P(V-V_0)=4.55\times 10^ {3}~\rm J$. So the heat added is $Q_{ac}= \Delta U _{ac}+W_{ac}=1.47\times 10^4 ~\rm J$. (c) The process $ca$ is where the heat is released. So $Q_{ac}$ is really the heat added to the engine. The engine efficiency is then the total work divided by $Q_{ac}$: $e= W/Q_{ac}=0.154$. (d) In a Carnot engine, $e_C = 1- T_c/T_h$. Since $T\propto PV$, $e_C = 1- P_0V_0/PV =0.75$. This is five times higher than the actual efficiency.

7-3

The initial volume: $V_i=50.0\times 10^{-6}~\rm m^3$. Initial pressure: $P_i=3.00\times 10^6 ~\rm Pa$. After adiabatic expansion (without loss of heat), the final volume is $V_f = 300\times 10^{-6} ~\rm m^3$. (a) From $P_iV_i^\gamma = P_fV_f^\gamma$, the unknown final pressure is $P_f = P_i (V_i/V_f)^\gamma = 2.44\times 10^5 ~\rm Pa$. (b) From the formula $W=\int_{V_i}^{V_f}P(V)dV$, where $P$ as a function of $V$ is $P(V)=V^{-\gamma}P_iV_i^\gamma$,

$$W==\int_{V_i}^{V_f} V^{-\gamma}P_iV_i^\gamma dV= {V_f^{1-\gamma}-V_i^{1-\gamma}\over 1-\gamma}P_iV_i^\gamma =192 ~\rm J~.$$

7-4

1 hp = 746 W. An Otto cycle has heat efficiency of $e=1- (V_1/V_2)^{\gamma -1}$. Here $V_1/V_2 = 1/6.20$ and $\gamma = 1.40$. Here the work done by the engine per second is $W=761~\rm J$, the heat absorbed each second is $Q_H = W(V_2/V_1)^{\gamma -1}=1580~\rm J$ and the heat exhausted is $Q_C = Q_H-W = 818 ~\rm J$ per second.

7-5

The coefficient of performance is $K=\vert Q_c\vert/\vert W\vert=\vert Q_C\vert/(\vert Q_H\vert-\vert Q_C\vert)= 5.7$. (a) Here $Q_C=42~\rm kcal$. So $Q_H = 49.4 ~\rm kcal$. (b) $W=\vert Q_H\vert-\vert Q_C\vert=7.4~\rm kcal$.