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An ideal gas is expanded from 1L to 6L i...

An ideal gas is expanded from 1L to 6L in a closed vessel at 2 atm pressure by applying heat at fixed temperature. Calculate work done and heat absorbed by the gas. [Given: 1 L atm=24.22 cal]

Text Solution

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Work done, `w=-P_(ex)DeltaV=-P_(ex)(V_(2)-V_(1))`
Given: `P_(ex)=2atm,V_(1)=1L and V_(2)=6L`
`therefore w=-2(6-1)=-10L*atm=-242.2cal`
`therefore`From the first law of thermodynamics, `DeltaU=q+w`.
Since the expansion is carried out isothermally and the system is an ideal gas, the change in internal energy in the process will be equal to zero. thus, `DeltaU=0`
`therefore 0=q+w or, q=-w=+242.2cal`
Therefore, amount of heat absorbed by the gas is 242.2 cal.
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