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Question

If the absolute temperature of an ideal gas having volume V $$cm^3$$ is doubled and the pressure is reduced to half, the final volume of gas will be?


A
0.25V
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B
0.50V
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C
2V2
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D
4V
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Solution

The correct option is D $$4V$$
Let $$T_1=T$$
$$\therefore T_2=2T$$

Let $$P_1=P$$
$$\therefore P_2=P/2$$

Given that, $$V_1=Vcm^3$$ 

$$P_1V_1=nRT_1$$

$$P_2V_2=nRT_2$$

$$\therefore \dfrac{P_1}{P_2}\times \dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$$

$$\therefore (2)\left(\dfrac{V}{V_2}\right)=\dfrac{1}{2}$$

$$\therefore V_2=4Vcm^3$$

Hence, option $$D$$ is correct.

Physics

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