The pressure of a gas changes linearli withe volume from 10kpa,200cc to 50kpa,50cc.

1)Calculate work done by gas

2)If no heat heat is supplied or extracred form gas,what is change in internal energy of gas?

when pressure changes linearly as given in question, if a graph is plotted acc. to the given data, then a line segment with a negative slope is obtained. neways, least bothered bout the slope.

so, i) Work done by the process is equal to area enclosed by the graph, i.e. area of a trapezium.

area = 1/2*base(here, the differense in volume)*sum of parallel sides(here, sum of pressures.)

A= 1/2*(200-150)*10^-6 *(10+50)*10³

A=4.5 J  (answer)

ii) no heat is supplied or extracted, i.e. dQ=0

we know that dQ=dU+dW. when dQ is 0, then dU= - dW, i.e. dU=-4.5 J

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