Question

Two Moles of an ideal gas are expanded isothermally & reversibly from 20L to 30 L at 300 K. Calculate the work done (R=8.314JK-1 mol-1).​

Answer 1

Given - Initial volume - 20 L

Final volume- 30 L

Temperature - 300 K

Number of moles- 2

R - 8.314 J/K/mol

To find- Work done, W

Solution- This problem can be easily solved by using the formula for work done in an isothermal and reversible process.

We know that

W = 2.303nRTlog (final vol/initial vol)

By putting the values we get

=[ 2.303× 2 × 8.314 × 300 × log ( 30/20)] J

= [114.88 × log ( 1.5) ]J

= [114.88 × 0.176 ] J

= 20.22 J

So, the work done is 20.22 J.

Answer 2

Answer:

By Formula-

Explanation:

W=2.303nRT log V2/V1

Put values n=2

R=8.314

T=300

V2=30 V1=20

Ans is -2023J

Hint-Log 1.5=0.176

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