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Thermodynamics

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Question : 5 of 10
Marks: +1, -0
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Solution:  
In first case, the state of the gas changes adiabatically from A to B.
Therefore, ΔQ=0,DW=−22.3 J\Delta Q = 0, \text{DW} = -22.3\ \mathrm{J}
If ΔU\Delta U is change in internal energy of the system then,
ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W
⇒0=ΔU+ΔW\Rightarrow 0 = \Delta U + \Delta W
ΔU=−ΔW=−(−22.3 J)\Delta U = -\Delta W = -(-22.3\ \mathrm{J})
=22.3 J= 22.3\ \mathrm{J}
In the second case, when the state A is taken to state B, the heat absorbed by the system
ΔQ=9.35 cal\Delta Q = 9.35\ \mathrm{cal}
=9.35×4.2 J=39.27 J= 9.35 \times 4.2\ \mathrm{J} = 39.27\ \mathrm{J}
=39.3 J,ΔW=?= 39.3\ \mathrm{J}, \Delta W = ?
Applying first law of thermodynamics
ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W
ΔW=ΔU−ΔQ\Delta W = \Delta U - \Delta Q
=39.3−22.3=17.0 J= 39.3 - 22.3 = 17.0\ \mathrm{J}
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