Question

A gas enclosed in a cylinder is expanded to double its initial volume (V¹=0.5 units)at a constant pressure of one atmosphere how much work is done in this pressure? is thi workdone on the gas or by the gas?how do you know this?​

Answer 1

Answer:

work done =-PdeltaV

P 1atm Vi=0.5 and final double V=1

work = -1(1-0.5)

work =-1

as work done is negative so work is done by the gas

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Answer 2

Answer:

The work is done by the gas which is 5.05 × $10^{4}$ J

Explanation:

Here we have been mentioned that the gas is enclosed in a cylinder where it has expanded to double its initial volume which was 0.5 units. This is done under constant pressure. Now to find the work done in this particular pressure we have the following formula:

W = P Δ V

Where,

W is Work done

P is Pressure

ΔV is a change in the volume  

Here we have, $V_{1} = 0.5 units$

$V_{2} =$ 2 × 0.5 units = 1 unit

P = 1 atm = 1.01 × $10^{5} Pa$

Putting them all in the formula we get work done to be:

∴ W = 1.01 × $10^{5}$ ( 1 - 0.5 ) J

⇒ W = 1.01 × $10^{5}$ × 0.5 J

⇒ W = 0.505 × $10^{5}$ J

⇒ W = 5.05 × $10^{4}$ J

Now here we can see that the work done is positive, and hence we deduce that the work is done by the gas to double its volume.