Answer:
-1215.9J is the work done
Explanation:
It is possible to find work done in the change of volume of a gas at constant pressure using:
W = -P×ΔV
Where W is work, P is pressure and V is change in volume}
Replacing:
W = -6atm×(5L-3L)
W = -12atmL
As 1atmL = 101,325J, work done in joules is:
-12atmL ×(101.325J / atmL) = -1215.9J is the work done
The value of work done when a volume increases from 3 liters to 5 liters at 6 atm of pressure is -1215.9 Joules.
Workdone on any boby or by the body will be calculated as:
W = -P×ΔV, where
W = workdone
P = applied or exerted pressure = 6 atm
ΔV = change in volume due to workdone = (5-3) L
Negative sign in the formula shows that work is done on the opposite side of the pressure or volume.
On putting all these values on the above equation, we get
W = -6atm × (5L-3L)
W = -12 atmL
We know that, 1 atmL = 101,325 J,
So, workdone in joules will be written as:
-12 atmL × (101.325J / atmL) = -1215.9 J
Hence, -1215.9 J is the workdone.
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