Question

Three moles of an ideal gas

are expanded isothermally from 15 dm3to 20 dm3

at constant external pressure of 1.2 bar. Estimate the amount of work in dm3bar and J.

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Answer 1

Solution :

W = - Pext ∆V = - Pext (V2 - V1)

Pext = 1.2 bar, V1= 15 dm3, V2 = 20 dm3

Substitution of these quantities into the

equation gives

W = -1.2 bar (20 dm3 - 15 dm3)

= -1.2 bar × 5dm3

= -6 dm3bar1 dm3

bar = 100 J

Hence, W = -6 dm3

bar × 100 J/dm3

bar = -600 J

Answer 2

Answer:

W = -6 dm³ bar  and -600 J

Explanation:

As per the given data,

3 moles of an ideal gas are expanded isothermally.

So initial volume V₁ = 15dm³

final volume V₂ = 20 dm³.

External pressure $P_{ext}$=1.2 bar

We know that W= $-P_{ext}$.ΔV

⇒W= $-P_{ext}$. (V₂-V₁)

⇒W= -1.2 bar.(20 dm³-15 dm³)

⇒W = -1.2 bar × 5 dm³

⇒W= - 6dm³ bar    

We know,

1 dm³ bar = 100 J

Thus W= -6 dm³· bar × $\frac{100J}{1 dm^{3} .bar} =$ -600 J.

Hence the amount of work is -6 dm³.bar and -600 J respectively.