Work for the following process ABCD on a monoatomic gas isA- W -2 P 0 V 01-ln 2B- W -2 P 0 V 0ln 2C- W-2 P0 V01-ln 2D- W -2 P 0 V 0ln 2

The correct option is D W= 2P_0V_0 ln2Work done W= W_AB+ W_BC+ W_CD nbsp; The process A— gt;B nbsp; and C— gt;D are isobaric where as the process B— ...

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Standard XII

Chemistry

Heat and Work

Work for the ...

Question

Work for the following process $A B C D$ on a monoatomic gas is

W = 2 P_{0} V_{0} l n 2

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W = - 2 P_{0} V_{0} (1 + l n 2)

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W = - 2 P_{0} V_{0} (1 + l n 2)

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W = - 2 P_{0} V_{0} l n 2

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Solution

The correct option is D. $W = - 2 P_{0} V_{0} l n 2$ .

The process A--->B and C--->D are isobaric where as the process B---->C is isothermal.
Work done along AB:
$W_{A B} = - P_{e x t} (V_{2} - V_{1}) = - P_{0} (2 V_{0} - V_{0}) = - P_{0} V_{0}$
Work done along BC:
$W_{B C} = - n R T l n (\frac{V_{2}}{V_{1}} = - n R T l n (\frac{4 V_{0}}{2 V_{0}}) = - n R T l n (2)$
Work done along CD:
$W_{C D} = - P_{e x t} (V_{2} - V_{1}) = - P (2 V_{0} - 4 V_{0}) = 2 P V_{0}$
as BC is an isothermal process, from B→C there is increase in volume by two times so the pressure will be half of original
So, $P = \frac{P_{0}}{2}$
$∴ W_{C D} = P_{0} V_{0}$
The total work done along the path $A B C D$ is
$W = W_{A B} + W_{B C} + W_{C D}$
$W = - P_{0} V_{0} - n R T l n (2) + P_{0} V_{0}$
$W = - P_{0} V_{0} - 2 P_{0} V_{0} l n (2) + P_{0} V_{0}$
(As the process BC is an isothermal process, then at point B, $2 P_{0} V_{0} = n R T)$
$W = - 2 P_{0} V_{0} l n 2$ .

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