Answer to Question #235647 in Molecular Physics | Thermodynamics for Rehema

Question

A system consisting of 1 kg of an ideal gas at 5 bar pressure and 0.02 m3

volume executes a cyclic process comprising the following three distinct operations : (i) Reversible

expansion to 0.08 m3 volume, 1.5 bar pressure, presuming pressure to be a linear function of

volume (p = a + bV), (ii) Reversible cooling at constant pressure and (iii) Reversible hyperbolic

compression according to law pV = constant. This brings the gas back to initial conditions.

(i) Sketch the cycle on p-V diagram.

(ii) Calculate the work done in each process starting whether it is done on or by the system

and evaluate the net cyclic work and heat transfer.

Answer 1

We have given data

$: m = 1 kg ; \\ p_{1} = 5 bar\\ V_{1} = 0.02 m^{3}\\ V_{2} = 0.08 m^{3}\\ p_{2} = 1.5 bar$

I)

(ii) Work done and heat transfer :

Process 1-2 (Linear law) :

p = a + bV  …(Given)

The values of constants a and b can be determined from the values of pressure and volume at the state points 1 and 2.

 We solve further

5 = a + 0.02b  …(i)

1.5 = a + 0.08b  …(ii)

By using (i) and (ii),

we get,

b = – 58.33 and a = 6.167

 So,

$W_{1-2}=\int_{1}^{2}pdV=\int_{1}^{2}\left ( a+bV \right )dV$

$=\int_{1}^{2}\left ( 6.167-58.33V \right )dV$

$=10^{5}\left | 6.167V-58.33\times \frac{V^{2}}{2} \right |_{0.02}^{0.08}$

$={10^{5}\left | 6.167\left ( 0.08-0.02 \right )-58.33\times \frac{\left ( 0.08^{2}-0.02^{2} \right )}{2} \right |\times 10^{-3}kJ} \\= 19.5$

This is work done by the system.

∙   Process 2 – 3 (constant pressure) :

 

$p_{3} = p_{2} = 1.5 p3 ​=p2 ​=1.5 bar$

 

$The \ volume \ V_{3}$

​ can be worked out from the hyperbolic compression 3–1, as follows :

 

$p_{1}V_{1}=p_{3}V_{3}$

​  So,  $V_{3}= \frac{p_{1}V_{1}}{p_{3}}=\frac{5\times 0.02}{1.5}=0.0667 m^{3}$

∴    $W_{2–3} = p_{2}(V_{3} – V_{2}) = 1.5 × 10^{5} (0.0667 – 0.08) × 10^{–3}$

= – 1.995 kJ

∙   Process 3 – 1 (hyperbolic process) :

$W_{3-1}=p_{3}V_{3}\log_{e}\left ( \frac{V_{1}}{V_{3}} \right )$

$V= (10^{5} × 1.5) × 0.0667 \log_{e}\left ( \frac{0.02}{0.0667} \right )\times 10^{-3} =(105 ×1.5)×0.0667loge ​(0.0667)$

 kJ = – 12.05 kJ.

This is the work done on the system.

 

Net work done,   $W_{net} = W_{1–2} + W_{2–3 }+ W_{3–1}$

​

 

= 19.5 + (– 1.995) + (– 12.05) = 5.445 kJ.

We know that the heat transferred during the complete cycle,$\oint \delta Q=\oint \delta W=5.455 ∮δQ=∮δW=\boxed{5.455 kJ}.$