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A vertical hollow cylinder contains an i...

A vertical hollow cylinder contains an ideal gas. The gas is enclosed by a `5kg` movable piston with an area of cross-section `5 xx 10^(-3)m^(2)`. Now, the gas is heated slowly from `300K` to `350K` and the piston rises by `0.1m`. The piston is now clamped at this position and the gas is cooled back to `300K`. Find the difference between the heat energy added during heating process and energy lost during the cooling process. `("1atm pressure" = 10^(5)Nm^(-2)]`

Text Solution

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Given, mass of piston M= 5kg, cross sectional area of piston `A =5xx10^(-3)m^2 , g = 10 ms^(-2)` and atmospheric pressure `P_0=10^5Nm^(-2)`
The initial pressure of the gas in the cylinder is
P= atmospheric pressure + pressure due to weight Mg of piston
`=P_0+(Mg)/A` or `P=10^5 +(5xx10)/(5xx10^(-6))=1.1xx10^5 Nm^(-2)`
When the piston rises by x = 0.1 m, the increase in the volume of the gas is
`DeltaV=Ax =(5xx10^(-3))xx0.1=5xx10^(-4)m^3`
Thus work done by the gas is
`W=int_(V_1)^(V_2)PdV`
or `=P(V_2-V_1)`
`=1.1xx10^5xx5xx10^(-4)`=55 J
If `DeltaU` is the increase in the internal energy during heating,then from the first law of thermodynamics, the heat energy supplied to the gas is given as
`Q=DeltaU+W=(DeltaU+55)` joule
Since the piston is clamped, the volume of the gas remains constant during cooling. Hence work done during cooling is zero.
As the gas is cooled back to the initial temperature, the change in the internal energy during cooling is
`DeltaU'=-DeltaU`
Thus heat energy lost by the gas during cooling is
`Q'=DeltaU'+W'=-DeltaU+0=-DeltaU`
The difference between heat energy added during heating and heat energy lost during cooling is
`DeltaQ=Q-Q'=(DeltaU+55)-DeltaU` = 55 joule .
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