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Question

The process $$CD$$ is shown in the diagram. As system is taken from $$C$$ to $$D$$, what happens to the temperature of the system?
807576_ee5191f1c75140529bbb97d75fd0ab4f.png


A
Temperature first decreases and then increases
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B
Temperatures first increases and then decreases
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C
Temperature decreases continuously
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D
Temperature increases continuously
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Solution

The correct option is A Temperature first decreases and then increases
Solution:- (A) temperature first decreases and then increases
From ideal gas equation,
$$PV = nRT$$
$$\Rightarrow T = \cfrac{PV}{nR}$$
Let $$P$$ be the mid point on $$CD$$.
At point C-
$$P = 3 {p}_{0}$$
$$V = {v}_{0}$$
$$\therefore {T}_{C} = \cfrac{3{p}_{0} {v}_{0}}{nR} ..... \left( 1 \right)$$
At mid-point P-
$$P = \cfrac{{p}_{0} + 3 {p}_{0}}{2} = 2 {p}_{0}$$
$$V = \cfrac{{v}_{0} + 3 {v}_{0}}{2} = 2 {v}_{0}$$
$$\therefore {T}_{P} = \cfrac{4 {p}_{0} {v}_{0}}{nR} ..... \left( 2 \right)$$
At point D-
$$P = {p}_{0}$$
$$V = 3 {v}_{0}$$
$$\therefore {T}_{D} = \cfrac{3 {p}_{0} {v}_{0}}{nR} ..... \left( 3 \right)$$
Now from $${eq}^{n} \left( 1 \right), \left( 2 \right) \& \left( 3 \right)$$, we have
$${T}_{C} < {T}_{P} > {T}_{D}$$
Hence the temperature will first decreases and then increases.

Chemistry

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