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  • One mole of an ideal monoatomic gas undergoes a thermodynamic cycle <img alt="\mathrm{1 \rightarrow 2 \rightarrow 3 \rightarrow 1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

One mole of an ideal monoatomic gas undergoes a thermodynamic cycle \mathrm{1 \rightarrow 2 \rightarrow 3 \rightarrow 1} as shown in the figure. Initial temperature of the gas is \mathrm{T_0=300 \mathrm{~K}}
process \mathrm{1-2: \mathrm{p}=\alpha \mathrm{V}}
process \mathrm{2-3: \mathrm{pV}= constant}
process \mathrm{3-1: p= constant.} The net work done by the cycle is:

 

Option: 1

2.8 \mathrm{RT}_0
 


Option: 2

3.8 \mathrm{RT}_0
 


Option: 3

4.8 \mathrm{RT}_0
 


Option: 4

5.8 \mathrm{RT}_0


Answers (1)

best_answer

\mathrm{ \mathrm{W}_{12}=\int_1^2 \mathrm{pdV}=\alpha \int_{\mathrm{V}_0}^{3 \mathrm{~V}_0} \mathrm{VdV}=\frac{\alpha}{2}\left[9 \mathrm{~V}_0^2-\mathrm{V}_0^2\right]=4 \alpha \mathrm{V}_0^2 }

Using gas law, \mathrm{ \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} }

\mathrm{ \text { or } \quad \mathrm{T}_2=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{P}_1 \mathrm{~V}_1} \mathrm{~T}_1=\frac{\mathrm{V}_2^2}{\mathrm{~V}_1^2} \mathrm{~T}_1=\left(\frac{3 \mathrm{~V}_0}{\mathrm{~V}_0}\right)^2 \mathrm{~T}_0=9 \mathrm{~T}_0 }

\mathrm{ \text { Now, } \quad \mathrm{W}_{23}=\mathrm{RT}_2 \ln \left|\frac{\mathrm{p}_2}{\mathrm{p}_3}\right|=\mathrm{R}\left(9 \mathrm{~T}_0\right) \ln \left|\frac{3 \mathrm{p}_0}{\mathrm{p}_0}\right|=9 R T_0 \ln |3|=9.8 R T_0 }

For isothermal process:\mathrm{ \mathrm{p}_2 \mathrm{~V}_2=\mathrm{p}_3 \mathrm{~V}_3}

Therefore,\mathrm{ \mathrm{V}_3=\frac{\mathrm{p}_2 \mathrm{~V}_2}{\mathrm{p}_3}=\frac{3 \mathrm{p}_0}{\mathrm{p}_0}\left(3 \mathrm{~V}_0\right)=9 \mathrm{~V}_0 }

Also,\mathrm{\mathrm{W}_{31} =\mathrm{p}_0\left(\mathrm{~V}_1-\mathrm{V}_3\right)=\mathrm{p}_0\left(\mathrm{~V}_0-9 \mathrm{~V}_0\right) }

\mathrm{ =-8 \mathrm{p}_0 \mathrm{~V}_0=-8 R \mathrm{~T}_0}

Applying gas law in the process \mathrm{1-2,}

\mathrm{ \mathrm{p}_0 \mathrm{~V}_0=\mathrm{RT}_0 \quad \text { or } \quad \alpha \mathrm{V}_0^2=\mathrm{RT}_0 }

\mathrm{\therefore \mathrm{W}_{12}=4 \alpha \mathrm{V}_0^2=4 \mathrm{RT}_0 }

The net work is

\mathrm{ \mathrm{W}_{\text {net }}=\mathrm{W}_{12}+\mathrm{W}_{23}+\mathrm{W}_{31}=4 \mathrm{RT}_0+9.8 \mathrm{RT}_0-8 \mathrm{RT}_0=5.8 \mathrm{RT_0} }


 

Posted by

Gunjita

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