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Question

1 litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

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Solution

Given:

γ = 1.5

T = 300 K

Initial volume of the gas, V_{1} = 1 L

Final volume, V_{2} = $\frac{1}{2}$ L

(a) The process is adiabatic because volume is suddenly changed; so, no heat exchange is allowed.

P_{1}V_{1}^{γ} = P_{2}V_{2}^{γ}

$\text{Or}{P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma}={P}_{1}(2{)}^{\gamma}\phantom{\rule{0ex}{0ex}}\frac{{P}_{2}}{{P}_{1}}={2}^{1.5}=2\sqrt{2}$

(b) P_{1} = 100 kPa = 10^{5} Pa

and P_{2} = $2\sqrt{2}$ × 10^{5} Pa

Work done by an adiabatic process,

$W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}\phantom{\rule{0ex}{0ex}}W=\frac{{10}^{5}\times {10}^{-3}-2\sqrt{2}\times {10}^{5}\times {\displaystyle \frac{1}{2}}\times {10}^{-3}}{1.5-1}\phantom{\rule{0ex}{0ex}}W=-82\mathrm{J}$

(c) Internal energy,

dQ = 0, as it is an adiabatic process.

⇒ dU = − dW = − (− 82 J) = 82 J

(d)

Also, for an adiabatic process,

T_{1}V_{1}^{γ−}^{1} = T_{2}V_{2}^{γ−}^{1}

T_{2} = T_{1}${\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma -1}$

= 300 $\times $ (2)^{0.5}

= 300 × $\sqrt{2}$ × = 300 × 1.4142

T_{2} = 424 K

(e) The pressure is kept constant.

The process is isobaric; so, work done = P$\u2206V$=nRdT.

Here, $n=\frac{PV}{RT}=\frac{{10}^{5}\times {10}^{-3}}{R\times 300}=\frac{1}{3R}$

So, work done = $\frac{1}{3R}\times R\mathit{\times}(300-424)=-41.4\mathrm{J}$

As pressure is constant,

$\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{2}}...\left(1\right)\phantom{\rule{0ex}{0ex}}{V}_{1}={V}_{2}\frac{{T}_{1}}{{T}_{2}}$

(f)Work done in an isothermal process,

$W=nRT\mathrm{ln}\frac{{V}_{2}}{{V}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3R}\times R\times T\times \mathrm{In}\left(2\right)$

= 100 × ln 2 = 100 × 1.039

= 103 J

(g) Net work done (using first law of thermodynamics)

= − 82 − 41.4 + 103

= − 20.4 J

γ = 1.5

T = 300 K

Initial volume of the gas, V

Final volume, V

(a) The process is adiabatic because volume is suddenly changed; so, no heat exchange is allowed.

P

$\text{Or}{P}_{2}={P}_{1}{\left(\frac{{V}_{1}}{{V}_{2}}\right)}^{\gamma}={P}_{1}(2{)}^{\gamma}\phantom{\rule{0ex}{0ex}}\frac{{P}_{2}}{{P}_{1}}={2}^{1.5}=2\sqrt{2}$

(b) P

and P

Work done by an adiabatic process,

$W=\frac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{\gamma -1}\phantom{\rule{0ex}{0ex}}W=\frac{{10}^{5}\times {10}^{-3}-2\sqrt{2}\times {10}^{5}\times {\displaystyle \frac{1}{2}}\times {10}^{-3}}{1.5-1}\phantom{\rule{0ex}{0ex}}W=-82\mathrm{J}$

(c) Internal energy,

dQ = 0, as it is an adiabatic process.

⇒ dU = − dW = − (− 82 J) = 82 J

(d)

Also, for an adiabatic process,

T

T

= 300 $\times $ (2)

= 300 × $\sqrt{2}$ × = 300 × 1.4142

T

(e) The pressure is kept constant.

The process is isobaric; so, work done = P$\u2206V$=nRdT.

Here, $n=\frac{PV}{RT}=\frac{{10}^{5}\times {10}^{-3}}{R\times 300}=\frac{1}{3R}$

So, work done = $\frac{1}{3R}\times R\mathit{\times}(300-424)=-41.4\mathrm{J}$

As pressure is constant,

$\frac{{V}_{1}}{{T}_{1}}=\frac{{V}_{2}}{{T}_{2}}...\left(1\right)\phantom{\rule{0ex}{0ex}}{V}_{1}={V}_{2}\frac{{T}_{1}}{{T}_{2}}$

(f)Work done in an isothermal process,

$W=nRT\mathrm{ln}\frac{{V}_{2}}{{V}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{1}{3R}\times R\times T\times \mathrm{In}\left(2\right)$

= 100 × ln 2 = 100 × 1.039

= 103 J

(g) Net work done (using first law of thermodynamics)

= − 82 − 41.4 + 103

= − 20.4 J

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