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Question

Starting with the same initial conditions, one mole of an ideal monoatomic gas expands reversibly from volume $$\displaystyle { V }_{ 1 }$$ to $$\displaystyle { V }_{ 2 }$$ in two different ways. The work done by gas is $$\displaystyle { w }_{ 1 }$$, if the process is purely isothermal, $$\displaystyle { w }_{ 2 }$$ if purely isobaric and $$\displaystyle { w }_{ 3 }$$ if purely adiabatic, then:


A
w2>w1>w3
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B
w2>w3>w1
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C
w1>w2>w3
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D
w1>w3>w2
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Solution

The correct option is A $$\displaystyle { w }_{ 2 }>{ w }_{ 1 }>{ w }_{ 3 }$$
W $$=-\int PdV$$
It is the area under p-V diagram

From the above image, we can say that the area under the P-V curve is in the increasing order: adiabatic, isothermal and isobaric

Hence, $$w_2>w_1>w_3$$

Chemistry

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