In changing the state of a gas adiabatically | vedclass

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Question

(A) For an adiabatic process,the heat exchange $\Delta Q = 0$. The work done on the system is $W_{adiabatic} = 22.3 \; J$. By the sign convention,work

Vedclass · Accepted Answer

$16.88$

Vedclass · Answer

(A) For an adiabatic process,the heat exchange $\Delta Q = 0$. The work done on the system is $W_{adiabatic} = 22.3 \; J$. By the sign convention,work done on the system is negative,so $\Delta W = -22.3 \; J$.From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,we get $0 = \Delta U - 22.3 \; J$,which implies $\Delta U = 22.3 \; J$.Since internal energy is a state function,$\Delta U$ remains the same for any process between states $A$ and $B$.In the second process,the heat absorbed is $\Delta Q' = 9.35 \; cal = 9.35 	imes 4.19 \; J = 39.1765 \; J$.Using the first law again: $\Delta Q' = \Delta U + \Delta W'$,where $\Delta W'$ is the work done by the system.$\Delta W' = \Delta Q' - \Delta U = 39.1765 \; J - 22.3 \; J = 16.8765 \; J$.Rounding to two decimal places,the work done by the system is $16.88 \; J$.

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