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(B) According to the $1^{	ext{st}}$ Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.Here,the gas releases heat,so $\Delta Q = -20 \, J$.Work i

Vedclass · Accepted Answer

$18$

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(B) According to the $1^{	ext{st}}$ Law of Thermodynamics,$\Delta Q = \Delta U + \Delta W$.Here,the gas releases heat,so $\Delta Q = -20 \, J$.Work is done on the gas,so $\Delta W = -8 \, J$.Substituting these values into the equation: $-20 \, J = \Delta U + (-8 \, J)$.$\Delta U = -20 \, J + 8 \, J = -12 \, J$.Since $\Delta U = U_{	ext{final}} - U_{	ext{initial}}$,we have $U_{	ext{final}} = U_{	ext{initial}} + \Delta U$.Given $U_{	ext{initial}} = 30 \, J$,then $U_{	ext{final}} = 30 \, J - 12 \, J = 18 \, J$.

In a thermodynamic process,the pressure of a fixed mass of a gas is changed in such a manner that the gas releases $20 \, J$ of heat when $8 \, J$ of work is done on the gas. If the initial internal energy of the gas was $30 \, J$,then the final internal energy will be ........ $J$.

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