Q.

In a thermodynamic process two moles of a monotomic ideal gas obeys pV2. If temperature of the gas increases from 300K to 400K, then find work done by the gas (where R= universal gas constant)

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a

-100R

b

200R

c

-20×0n

answer is B.

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Detailed Solution

pV2= constant

(as pV-2 )

Comparing with pVx= constant, we have

x=2

W=nRΔT1-x

Substituting the values, we have

W=(2)(R)(400-300)1-2

=-200R

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In a thermodynamic process two moles of a monotomic ideal gas obeys p∝V2. If temperature of the gas increases from 300K to 400K, then find work done by the gas (where R= universal gas constant)