How do I solve this gas problem?

Dec 9, 2017

Explained below ...

Explanation:

[1] Ideal Gas EoS: \quad PV=nRT; (always true)

[2] Monoatomic ideal gases: If $U$ is the internal energy, ${C}_{p}$ and ${C}_{v}$ are molar heat capacities at constant volume and constant pressure, respectively,

$U = \frac{3}{2} n R T$

C_p = 5/2R;\qquad C_v=3/2R;\qquad \gamma=C_p/C_v=5/3;

EoS: $\setminus \quad P {V}^{\setminus} \gamma = c o n s t .$ (true only for adiabatic process)

Work Done: ${W}_{i \setminus \rightarrow f} = \frac{{p}_{f} {V}_{f} - {p}_{i} {V}_{i}}{\setminus \gamma - 1}$

Problem
The three intermediate points $a$, $b$ and $c$ are characterized by the state variables (p_a, V_a); \quad(p_b, V_b) and $\left({p}_{c} , {V}_{c}\right)$.

Observe that: $\setminus \quad {V}_{a} = {V}_{b}$ and $\setminus \quad {p}_{a} = {p}_{c}$

Step 1 (Adiabatic Expansion): During adiabatic process the system does not exchange heat with the surroundings.

Q_{bc}=0;
p_bV_b^\gamma = p_cV_c^\gamma; \qquad \rightarrow p_c/p_b = (V_b/V_c)^\gamma

Remember that $\setminus \quad {p}_{a} = {p}_{c}$ and ${V}_{c} = 8 {V}_{b} = 8 {V}_{a}$,
${p}_{c} / {p}_{b} = \frac{1}{8} ^ \setminus \gamma = {p}_{a} / {p}_{b}$......(Eq1)

Next calculate the work done -

${W}_{b c} = \frac{{p}_{c} {V}_{c} - {p}_{b} {V}_{b}}{\setminus \gamma - 1} = \frac{{p}_{b} {V}_{b}}{\setminus \gamma - 1} \left[\frac{{p}_{c} {V}_{c}}{{p}_{b} {V}_{b}} - 1\right]$
${W}_{b c} = - \frac{{p}_{b} {V}_{b}}{\setminus \gamma - 1} \left[1 - \frac{1}{8} ^ \left(\setminus \gamma - 1\right)\right]$......(Eq2)

Work done is negative, as is expected for expansion. Work is being done by the gas.

Step 2 (Isobaric Compression): For isobaric processes the ratio of volume to temperature is a constant (Charles's law).

V_c/T_c = V_a/T_a; \qquad \rightarrow T_c/T_a = V_c/V_a=8

Change in internal energy -
$\setminus \Delta {U}_{c a} = \frac{3}{2} n R \setminus \Delta {T}_{c a} = \frac{3}{2} n R \left({T}_{a} - {T}_{c}\right) = \frac{3}{2} n R {T}_{a} \left(1 - {T}_{c} / {T}_{a}\right)$
$\setminus \Delta {U}_{c a} = \frac{3}{2} n R {T}_{a} \left(1 - 8\right) = - \frac{21}{2} \left(n R {T}_{a}\right)$

Using the ideal gas equation of state, $\setminus \quad {p}_{a} {V}_{a} = n R {T}_{a}$ and recognizing that ${V}_{a} = {V}_{b}$,

$\setminus \Delta {U}_{c a} = - \frac{21}{2} {p}_{a} {V}_{a} = - \frac{21}{2} {p}_{a} {V}_{b}$

Using Eq1 : $\setminus \quad {p}_{a} = {p}_{b} / {8}^{\setminus} \gamma$
$\setminus \Delta {U}_{c a} = - \frac{21}{2} \left({p}_{b} / {8}^{\setminus} \gamma\right) {V}_{b} = - \frac{21}{2} \frac{{p}_{b} {V}_{b}}{8} ^ \setminus \gamma$......(Eq3)

Work done during an isobaric process is-

${W}_{c a} = - {p}_{a} \setminus \Delta {V}_{c a} = - {p}_{a} \left({V}_{a} - {V}_{c}\right) = {p}_{a} \left({V}_{c} - {V}_{a}\right)$
${W}_{c a} = {p}_{a} {V}_{a} \left({V}_{c} / {V}_{a} - 1\right) = {p}_{a} {V}_{a} \left(8 - 1\right) = 7 {p}_{a} {V}_{a}$

Using Eq1: $\setminus \quad {p}_{a} = {p}_{b} / {8}^{\setminus} \gamma$ and the fact that ${V}_{a} = {V}_{b}$,

${W}_{c a} = 7 \left({p}_{b} / {8}^{\setminus} \gamma\right) {V}_{b} = 7 \frac{{p}_{b} {V}_{b}}{8} ^ \setminus \gamma$......(Eq4)

Work done is positive, as it is expected for compression. Work is being done on the gas.

Next apply the first law of thermodynamics to calculate the heat,

\Delta U_{ca}=Q_{ca}+W_{ca}; \qquad Q_{ca}=\DeltaU_{ca}-W_{ca}

${Q}_{c a} = - \left(\frac{21}{2} + 7\right) \frac{{p}_{b} {V}_{b}}{8} ^ \setminus \gamma = - \frac{35}{2 \setminus \times {8}^{\setminus} \gamma} {p}_{b} {V}_{b}$ ......(Eq 5)

Negative value for ${Q}_{c a}$ indicates that heat leaves the system.

Step 3 (Isochoric Heating): For an isochoric process volume is a constant and the ratio of pressure to temperature is a constant. Since the volume of the gas does not change, the work done during an isochoric process is zero.

W_{ab}=0;

p_a/T_a = p_b/T_b; \qquad \rightarrow T_a/T_b = p_a/p_b=1/8^\gamma;

The temperature the gases raises from ${T}_{a}$ to ${T}_{b}$. So heat must have entered the system. Use the temperature difference and molar heat capacity for constant volume process to find this heat

${Q}_{a b} = n {C}_{v} \setminus {\Delta}_{a b} = n {C}_{v} \left({T}_{b} - {T}_{a}\right) = n {C}_{v} {T}_{b} \left(1 - {T}_{a} / {T}_{b}\right)$
${Q}_{a b} = n \left(\frac{3}{2} R\right) {T}_{b} \left(1 - \frac{1}{8} ^ \setminus \gamma\right) = \frac{3}{2} \left(n R {T}_{b}\right) \left(1 - \frac{1}{8} ^ \setminus \gamma\right)$

Using the ideal gas EoS : $\setminus \quad {p}_{b} {V}_{b} = n R {T}_{b}$,

${Q}_{a b} = \frac{3}{2} \left(1 - \frac{1}{8} ^ \setminus \gamma\right) {p}_{b} {V}_{b}$ ...... (Eq 6)

As expected, ${Q}_{a b}$ is positive, indicating that heat is added to the system.

Numerical Calculation:
${p}_{b} = 12 \setminus \quad a t m = \left(12 \setminus \quad \cancel{a t m}\right) \setminus \times \left(1.01325 \setminus \times {10}^{5} \setminus \quad \frac{P a}{\cancel{a t m}}\right)$
p_b = 1.21\times10^6\quad N/m^2; \qquad V_b = 3.60\times10^{-3}\quadm^3;

${p}_{b} {V}_{b} = 4.35 \setminus \times {10}^{3} \setminus \quad N . m = 4.35 \setminus \quad k J$.
$\setminus \gamma = \frac{5}{3}$

(a) Heat added to the gas:
Q_{ab}=3/2(1-1/8^\gamma)p_bV_b = +6.32\quad kJ;

(b) Heat leaving the gas:
Q_{ca}=-35/(2\times8^\gamma)p_bV_b = -2.38\quad kJ;

(c) Net Work Done:
W_{bc}=-(p_bV_b)/(\gamma-1)[1-1/8^{\gamma-1}]=-4.89\quad kJ;
W_{ca}=+0.95\quad kJ;

${W}_{\text{net}} = {W}_{b c} + {W}_{c a} = - 3.94 \setminus \quad k J$
Negative sign is because the work is done by the gas. Negative sign is as a result of sign convention, according to which the work done on the gas is positive and the work done by the gas is negative.

(d) Efficiency of the cycle:
$\setminus \eta \setminus \equiv | W \frac{|}{Q} _ \left\{\text{in"}=(|Q_{"in"}|-|Q_{"out"}|)/Q_{"in"}=1-|Q_{"out"}|/Q_{"in}\right\} = 1 - | {Q}_{c a} \frac{|}{Q} _ \left\{a b\right\}$

\eta=62.3%