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Q.

Two moles of an ideal monoatomic gas is taken through a cyclic process as shown in the P-T diagram. In the process BC, $P T^{- 2} =$ constant. Then the ratio of heat absorbed and heat released by the gas during the process AB and process BC respectively is

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a

2

b

6

c

3

d

5

answer is C.

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Detailed Solution

$Δ Q_{A B} = n C p Δ T = 2 \times \frac{5 R}{2} (2 T_{0} - T_{0}) = 5 R T_{0}$

In the process BC, $P T^{- 2} = c o n s \tan t$
$p p^{- 2} V^{- 2} =$ constant
$P V^{2} =$ constant
$∴$ molar heat Capacity
$c = c_{v} + \frac{R}{1 - x} = \frac{3 R}{2} + \frac{R}{1 - 2} c = \frac{R}{2} ∴ Δ Q_{B C} = n C Δ T = 2 \times \frac{R}{2} (T_{0} - 2 T_{0}) = - R T_{0} ∴ | \frac{Δ Q_{A B}}{Δ Q_{B C}} | = \frac{5 R T}{R T_{0}} = 5$

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Two moles of an ideal monoatomic gas is taken through a cyclic process as shown in the P-T diagram. In the process BC, PT−2= constant. Then the ratio of heat absorbed and heat released by the gas during the process AB and process BC respectively is