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

Grade:12

## 1 Answers

Ashwani Dubey
17 Points
2 years ago
Solution:- The temperature will decrease first the will increase.
From ideal gas law,
$PV = nRT \\ \Rightarrow T = \frac{PV}{nR} \\ \text{Let P be the mid point on CD}. \\ \text{At point C}- \\ P = 3 {p}_{0} \\ V = {v}_{0} \\ \therefore {T}_{C} = \cfrac{3{p}_{0} {v}_{0}}{nR} ..... \left( 1 \right) \\ \text{At mid-point P}- \\ P = \cfrac{{p}_{0} + 3 {p}_{0}}{2} = 2 {p}_{0} \\ V = \cfrac{{v}_{0} + 3 {v}_{0}}{2} = 2 {v}_{0} \\ \therefore {T}_{P} = \cfrac{4 {p}_{0} {v}_{0}}{nR} ..... \left( 2 \right) \\ \text{At point D}- \\ P = {p}_{0} \\ V = 3 {v}_{0} \\ \therefore {T}_{D} = \cfrac{3 {p}_{0} {v}_{0}}{nR} ..... \left( 3 \right) \\ \text{Now from } {eq}^{n} \left( 1 \right), \left( 2 \right) \& \left( 3 \right),\text{ we have} \\ {T}_{C} < {T}_{P} > {T}_{D} \\$
Hence the temperature will first decrease and then increases.

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