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Standard XII

Physics

Work

A cylinder of...

Question

A cylinder of ideal gas is closed by an 8 kg movable piston $(a r e a 60 c m^{2})$ as shown in fig. atmospheric pressure is $100 k P a$ . When the gas is heated from $30^{o} C$ to $100^{o} C$ the piston rises by $20 c m$ . The piston is then fixed in its place and the gas is cooled back to $30^{o} C$ . Let $△ Q_{1}$ be the heat added to the gas in the heating process and $| △ Q_{2} |$ the heat lost during cooling . then the value of $[△ Q_{2} - | △ Q_{2} |]$ will be :

Zero

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136 J

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- 136 J

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- 68 J

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Solution

The correct option is B $136 J$
$Q_{1} = d U + d W = n c_{v} d T + P d W = n c_{v} (70) + P A x$
x=0.2m
When the piston is fixed dV=0 therefore workdone is zero
$Q_{2} = d U = n c_{v} d T = n c_{v} (70)$

Difference of heat= $P A x = (P_{a t m} + m g / A) A x = 136 J$

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The correct option is B 136JQ1=dU+dW=ncvdT+PdW=ncv(70)+PAxx=0.2mWhen the piston is fixed dV=0 therefore workdone is zeroQ2=dU=ncvdT=ncv(70)Difference of heat= PAx=(Patm+mg/A)Ax=136J

The correct option is B $136 J$
$Q_{1} = d U + d W = n c_{v} d T + P d W = n c_{v} (70) + P A x$
x=0.2m
When the piston is fixed dV=0 therefore workdone is zero
$Q_{2} = d U = n c_{v} d T = n c_{v} (70)$

Difference of heat= $P A x = (P_{a t m} + m g / A) A x = 136 J$