Q

# 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 $\dpi{100} ABCDA$, as shown in the $\dpi{100} P-T$ diagram.

Question: The net work done on the gas in the cycle   $\dpi{100} ABCDA$  is

• Option 1)

zero

• Option 2)

$276R$

• Option 3)

$1076R$

• Option 4)

$1904R$

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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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