n moles of an ideal gas undergo an isothermal process at temperature T. P–V graph of the process is as shown in the figure. A point $A (V_{1}, P_{1})$ is located on the P–V curve. Tangent at point A, cuts the V–axis at point D. AO is the line joining the point A to the origin O of PV diagram. Then,

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Coordinates of points D is $(\frac{3 V_{1}}{2}, 0)$

Coordinates of points D is $(2 V_{1}, 0)$

Area of the triangle AOD is $\frac{3}{4} n R T$

Area of the triangle AOD is $n R T$

answer is B, C.

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Detailed Solution

$\tan α = - [s l o p e a t (P_{1}, V_{1})] = \frac{P_{1}}{V_{1}}$

$\tan θ = \frac{P_{1}}{V_{1}}$

$∴ α = θ$

$O M = M D = V_{1}$

$∴ O D = 2 V_{1}$

Area of triangle $A O D = \frac{1}{2} (A M \times O D) = \frac{1}{2} (P_{1} \times 2 V_{1}) = P_{1} V_{1} = n R T$

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