Question

derive an expression for work done by the gas in an isothermal process.​

Answer 1

000009999990999999900000000000

Answer 2

Answer:

Let us consider 1 mole of gas is enclosed in an isothermal container.

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Let P1,P1,V1V1 and T be the initial pressure, initial volume, and temperature. As work is done, let the gas expand to P2,P2,V2V2 where P2P2 is the reduced pressure and V2V2 is the expanded volume.

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Since the process is an Isothermal Process, the temperature remains constant. We know that work done is given by,

W=∫dWW=∫dW

⇒W=∫V2V1PdV________(1)⇒W=∫V1V2PdV________(1)

We have the relation PV=nRTPV=nRT

⇒PV=RT⇒PV=RT (∵n=1mole)(∵n=1mole) and R is the ideal gas constant.

⇒P=RTV⇒P=RTV

Substituting the value of P in equation 1,

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we get

⇒W=RT∫V2V1dVV⇒W=RT∫V1V2dVV

⇒W=RT[lnV]V2V1⇒W=RT[ln⁡V]V1V2

⇒W=RT[lnV2−lnV1]⇒W=RT[ln⁡V2−ln⁡V1]

⇒W=RTlnV2V1⇒W=RTln⁡V2V1

∴W=2.303RTlog10V2V1∴W=2.303RTlog10V2V1

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We know that for constant temperature,

P1P2=V2V1P1P2=V2V1

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Thus, W=2.303RTlog10P1P2W=2.303RTlog10P1P2

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Thus, work done by the gas in an isothermal process is given by the expression, W=2.303RTlog10P1P2W=2.303RTlog10P1P2.

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Happy learning ☺!