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Question

On a TP diagram, two moles of ideal gas perform process AB and CD. If the work done by the gas in the process AB is two times the work done in the process CD then what is the value of $$T_1/ T_2$$?

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A
12
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B
1
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C
2
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D
4
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Solution

The correct option is D $$\dfrac{1}{2}$$


Work done in Isothermal process 
$$ W=nRT\ell n\cfrac { { P }_{ 1 } }{ { P }_{ 2 } } \longrightarrow \left( 1 \right) $$
$$\therefore W_{AB}=2RT_1 \ell n\cfrac{P_A}{P_B}$$
$$\Rightarrow  W_{CD}=2RT_2 \ell n\cfrac{P_C}{P_D}$$
$$T$$ varies proportionally to $$P$$ 
$$\cfrac{P_A}{P_B}=\cfrac{P_C}{P_D}\longrightarrow \left( 2 \right) $$
$$\Rightarrow  W_{AB}=2W_{CD}$$
$$\Rightarrow  \cfrac{W_{AB}}{W_{CD}}=\cfrac{T_1}{T_2}=\left( \cfrac { 1 }{ 2 }  \right) ^{ -1 }$$
$$\Rightarrow  \cfrac{T_2}{T_1}=\cfrac{1}{2}.$$
Hence, the answer is $$\cfrac{1}{2}.$$


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Physics

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