Question

Find the work done when 2 moles of hydrogen expand isothermally from 15 to 50 litres against a constant pressure of 1 atm at 25 degree celsius

Answer 1

W = –P (V2 – V1) 
= – 1 × 35 = – 35 L-atm 
= – 35 × 101.3 J = –3545.5 J 
= – 848.2 cal

Answer 2

Given:

P = 1 atm

$V_1$ = 15 L

$V_2$ = 50 L

To find:

Work done = ?

Formula to be used:

Work done = - P($V_2-V_1$)

Calculation:

Work done = - P($V_2-V_1$)

Work done = - 1(50-15)

Work done = - 1 × 35 = -35 l/atm

As 1 l/atm = 101.3 J,

∴ -35 l/atm = -35×101.3 = -3545.5J.

As 1 calorie = 4.184 J

∴ -3545.5 J = $\frac{-3545.5 J}{4.184J} = -848.2 cals$

Conclusion:

Thus the work done is identified as -848.2cals.

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