Homework Assignment # 6

Due Wed Oct 23, 2019

Consider the transformations shown in Fig. 6.3 (reproduced just below) and discussed in class

The following table lists the initial and final pressures, temperatures, and volumes, and well as $\Delta U,\,w,\,q$, and $\Delta S$ for the five reversible transformations shown in the figure above. One mole of a monatomic ideal gas is assumed. The directions of each transformation are indicated in the first column.
step $P_i$ $V_i$ $T_i$ $P_f$ $V_f$ $T_f$ $\Delta U$ $w_{rev}$ $q_{rev}$ $\Delta S =\int \frac{\delta q_{rev}}{T}$
A (isothermal)
$V_1\to V_2$		 $P_1$ $V_1$ $T_1$ $\frac{1}{2}P_1$ $2 V_1$ $T_1$ 0 $-RT_1 \ln 2$ $+RT_1 \ln 2$ $R \ln 2$
B (adiabatic)
$V_1\to V_2$		 $P_1$ $V_1$ $T_1$ $2^{-5/3}P_1$ $2V_1$ $\frac{1}{2}^{2/3}T_1$

$=0.630 T_1$

 $\frac{3}{2}R\left [\left ( \frac{1}{2}\right )^{2/3}-1\right]T_1$

$=-0.555R T_1$

 $-0.555 RT_1$ 0 0
C (isochoric)
$T_1\to T_2$		 $\frac{1}{2}P_1$ $2 V_1$ $T_1$ $2^{-5/3}P_1$ $2V_1$ $T_f=T_2$

$=0.630 T_1$

 $\frac{3}{2}R (T_2-T_1)$

$=-0.555 R T_1$

 0 $-0.555 R T_1$ $\frac{3}{2} R \int \frac{dT}{T}$

$=\frac{3}{2}R \ln(\frac{1}{2}^{2/3})$

$= - R \ln 2$
D (isobaric)
$V_1\to V_2$		 $P_1$ $V_1$ $T_1$ $P_1$ $2 V_1$ $2 T_1$ $\frac{3}{2} R T_1$ $-P_1 V_1$

$= -R T_1$

 $\frac{5}{2}R T_1$ $R(\frac{3}{2}\int \frac{dt}{T}+ \int \frac{dv}{V})$
$=\frac{5}{2}R \ln 2$
E (isochoric)
$T_1\to T_3$		 $\frac{1}{2}P_1$ $2 V_1$ $T_1$ $P_1$ $2 V_1$ $2T_1$ $\frac{3}{2}R T_1$ 0 $\frac{3}{2}R T_1$ $\frac{3}{2}R \ln 2$
  1. Verify all the entries for steps B and C (we did A, D, and E in class).
  2. Consider an isobaric transition that goes from $P_3,V_2$ to $V_1$. What will be the values of $P_f$ and $T_f$ for this transition? At constant pressure Pf=Pi=P3. Also, at constant pressure the ratio T/V is constant, so Tf= Ti (Vf / Vi). Here, since V2=2V1, the temperature will be halved as we compress the gas at constant pressure.
  3. Derive $\Delta U,\,w,\,q$ and $\Delta S$ for (all processes are reversible)
    1. An isochoric transition that goes from $P_3,V_2,T_2$ to $P_1,V_2,T_3$. For an isochoric transition: $\delta w=0$, so $\Delta U = C_V \Delta T = ({\rm here})\frac{3}{2} (T_3-T_2)$

      isochoric, so $dV=0$, so $w=0$.

      $q=C_V \Delta T = \frac{3}{2}T (T_3-T_2)= ({\rm see\,above\,table}) \frac{3}{2}T (2 T_1 - 0.630 T_1) = 2.055 R$. The units are (J/K $\times$ K) = J.

      $dS = dq_{rev}/T = C_V dT/T$ so $\Delta S = \frac{3}{2} R \int dT/T = \frac{3}{2} R ln(T_3/T_2) =\frac{3}{2} R ln (2 T_1 / 0.630 T_1) = \frac{3}{2} R ln (3.175) = \frac{3}{2} \times 1.16 R$

    2. The isobaric transition defined in question (2) above.
      For an isobaric transition,

      $w=-\int PdV = P(const) \Delta V= -P_3(V_1-V_2)= P_3 V_1$ (positive)

      $dH=C_P dT +(dH/dP)_T dP$. But $dP=0$ for isobaric, so $\Delta H = \frac{5}{2}R(T_f-T_i)$. Now, since $H=U+PV$, $\Delta H = \Delta U + P\Delta V + V\Delta P$. The last term vanishes since $P$ is constant. Thus
      $\Delta U = \Delta H - P\Delta V = (5/2)R(T_f-T_i)- P_3 (V_f-V_i)$

      $q = \Delta U - w = \frac{5}{2}R(T_f-T_i)- P_3 (V_f-V_i)+ P_3(V_f-V_i)=\frac{5}{2}R(T_f-T_i)$ (We could also solve this by remembering that $C_P=\frac{5}{2}R$ is the heat capacity at constant pressure, so $q_P=\frac{5}{2}R \Delta T$

      Now, $dS =(\delta q_{rev}/T) dT = C_P dT/T$. So, $\Delta S = C_P ln(T_f/T_i) = C_P \ln(1/2) = -\frac{5}{2} R \times 0.693$

    3. The cyclic process $P_1,V_1,T_1\to P_1,V_2,T_3\to P_3,V_2,T_2 \to P_1,V_1,T_1$.
      Referencing the table at the beginning of the problem set, the cyclic process is D--E+C--B. Adding up the corresponding entries in the table gives

      $\delta U = \frac{3}{2}R T_1 -\frac{3}{2}R T_1 -0.555RT_1 - (-0.555RT_1) = 0$, which we expect.

      $w = -RT_1-0+0-(-0.555RT_1) = -0.445 RT_1$

      $q = \Delta U - w = +0.445 RT_1$

      $\Delta S = \frac{5}{2}R \ln 2-\frac{3}{2}R\ln 2- R \ln 2 -0 = 0$ (which we expect)