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I have a doubt, kindly clarify. - Chemical Thermodynamics - JEE Main-3

Directions: Question are based on the following paragraph.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram.

Question: The net work done on the gas in the cycle   ABCDA  is

  • Option 1)

    zero

  • Option 2)

    276R

  • Option 3)

    1076R

  • Option 4)

    1904R

 
Answers (1)
135 Views
V Vakul

As we learnt in @

Reversible Isobaric Process -

W=P\left ( V_{f}-V_{i} \right )
 

- wherein

P_{gas}= constant

Since pressure of gas is defined during whole of the process and hence the process must be reversible

 

 and

 

Work Done for Isothermal Reversible Compression of an ideal gas -

W= -nRT\ln \frac{V_{f}}{V_{i}}

V_{i}> V_{f}

 

- wherein

At constant temperature

P\, \propto \frac{1}{V}

W= -nRT\, \ln \frac{P_{i}}{P_{f}}

P_{f}> P_{i}

W_{AB}=-2\times R\times 200 = -400 R

W_{BC}=+500R\ ln2

W_{CD}=-2\times R\times (-200)=+400R

W_{DA}=+R\times 300\ln \frac{1}{2}=-300R\ln2

W_{ABCDA}=200R \times ln2=276R

 

 

 

 


Option 1)

zero

This option is incorrect.

Option 2)

276R

This option is correct.

Option 3)

1076R

This option is incorrect.

Option 4)

1904R

This option is incorrect.

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