At $25^{o} C$ , a $0.01$ mole sample of a gas is compressed in volume from $4.0 L$ to $1.0 L$ at constant temperature. What is work done for this process if the external pressure is $4.0$ bar?

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Solution

External pressure $(P) = 4 bar = 3.95 atm$
Initial volume $(V_{1}) = 4 L$
Final volume $(V_{2}) = 1 L$
No. of moles $(n) = 0.001 mole$
$R = 8.314 L - a t m K^{- 1} {m o l}^{- 1}$

For an isothermal process,
Work done $(W) = - P Δ V$

$∴ W = - 3.95 \times (1 - 4) = 11.85 L . a t m = 1200.7605 = 1.2 \times 10^{3} J$

Hence the work done for the given process is $1.2 \times 10^{3} J$ .

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At 25oC, a 0.01 mole sample of a gas is compressed in volume from 4.0L to 1.0L at constant temperature. What is work done for this process if the external pressure is 4.0 bar?

External pressure (P)=4 bar=3.95 atmInitial volume (V1)=4LFinal volume (V2)=1LNo. of moles (n)=0.001 moleR=8.314L−atmK−1mol−1For an isothermal process,Work done (W)=−PΔV∴W=−3.95×(1−4)=11.85L.atm=1200.7605=1.2×103JHence the work done for the given process is 1.2×103J.

At $25^{o} C$ , a $0.01$ mole sample of a gas is compressed in volume from $4.0 L$ to $1.0 L$ at constant temperature. What is work done for this process if the external pressure is $4.0$ bar?