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Q.

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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a

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

b

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

c

area of the triangle AOD is $n R T$

d

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

answer is B, C.

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Detailed Solution

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Sol. $\tan α = - [s l o p e {at (P}_{1}, V_{1})] = \frac{P_{1}}{V_{1}}$
$\tan θ = \frac{P_{1}}{V_{1}} ∴ α = θ O M = M D = V_{1}$
∴ $O D = 2 V_{1}$
Area of triangle AOD
$= \frac{1}{2} (A M \times O D) = \frac{1}{2} (P_{1} \times 2 V_{1}) = P_{1} V_{1} = n R T$

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