Chemistry, asked by mirfayzanfazu642, 3 months ago

write the expression for reversible work done.​

Answers

Answered by sanskarsingh98013416
3

Answer:

If the external pressure is zero, the work done is also zero as the gas expands into the vacuum. ... In the reversible process, Pext is always less than the pressure of the gas, by an infinitesimally small quantity. W = (P – dp) dV. In the equation W tends to the maximum as (P – dp) tends to P or dp tends to zero.

Answered by simran5972
0

Answer:

ANSWER

Solution:-

Let us consider n moles of ideal gas enclosed in a cylinder fitted with a weightless and frictionless piston. The work of expansion for a small change of volume dV against the external pressure P is given by-dW=−PdV

∴ Total work done when the gas expands from initial volume V

1

to the final volume V

2

, will be-

W=−∫

V

1

V

2

PdV

For an ideal gas,

PV=nRT

⇒P=

V

nRT

Therefore,

W=−∫

V

1

V

2

V

nRT

dV

For isothermal expansion, T is constant.

Therefore,

W=−nRT∫

V

1

V

2

V

dV

⇒W=−nRT[lnV]

V

1

V

2

⇒W=−nRT(lnV

2

−lnV

1

)

⇒W=−nRTln(

V

1

V

2

)

⇒W=−2.303nRTlog(

V

1

V

2

).....(1)

At constant temperature,

P

1

V

1

=P

2

V

2

V

1

V

2

=

P

2

P

1

Therefore,

W=−2.303nRTlog(

P

2

P

1

).....(2)

Equation (1)&(2) are the expression for the work obtained in an isothermal reversible expansion of an ideal gas.

Explanation:

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