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Question

$$P-T$$ diagram of one mole of an ideal monoatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is
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A
2.5RT
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B
2RT
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C
1.5RT
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D
3.5RT
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Solution

The correct option is B $$2.5 RT$$
let $$A=1,B=2,C=3,D=4$$
now $${P}_{1}={P}_{4}={P}$$ & $${P}_{2}={P}_{3}={2P}$$

so by ideal gas equation we have
$${V}_{1}$$=$$\dfrac{3RT}{P}$$    $${V}_{2}$$=$$\dfrac{RT}{2P}$$

now,
1) for process 1-2
$${W}_{1-2}$$=$$\dfrac{{P}_{1}{V}_{1}-{P}_{2}{V}_{2}}{{\gamma}-{1}}$$
so $${ W }_{ 1-2 }=\dfrac { 2P\dfrac { RT }{ 2P } -P\dfrac { 3RT }{ P }  }{ 0.4 } =-5RT$$

2)for process 3-4
$${ W }_{ 3-4 }=\dfrac { 2P\dfrac { 2RT }{ P } -P\dfrac { 5RT }{ P }  }{ 0.4 } =-2.5RT$$

3) for process 4-1
$${ W }_{ 4-1 }={ P }_{ 1 }({ V }_{ 4 }-{ V }_{ 1 })$$
$${ W }_{ 4-1 }={ P }[\dfrac { 5RT }{ P } -\dfrac { 3RT }{ P } ]=2RT$$

4) for process 2-3
$${ W }_{ 2-3 }={2 P }[\dfrac { 2RT }{ P } -\dfrac { RT }{2 P } ]=3RT$$

now adding[1,2,3,4] we get the net work done
$${W}_{total}$$=$${W}_{1-2}+{W}_{2-3}+{W}_{3-4}+{W}_{4-1}$$
=$${-5RT}+{3RT}+{-2.5RT}+{2RT}={-2.5RT}$$

Physics

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