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A student is calculating the work durin...

A student is calculating the work during a reversible isothermal process , show by 2 mole of an ideal
gas . He by mistake calculate the area as show in the PV graph (Shaded area ) equal to 49.26 liter atm .
Calculate the correct value of work ( in litre atm ) during the process.
(Given : R =0.0821 liter atm // "mole"/K)

A

`49.26`

B

`-34.14`

C

`-78.63`

D

`-98.52`

Text Solution

Verified by Experts

For isothremal process
`W= -" nRT In " (V_(2))/(V_(1))`
` = - 49.26 xx "In" 2`
` =- 34. 14 " "["Shaded area = PV = nRT"]`
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