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Standard XII

Physics

Thermodynamic Processes

Two moles of ...

Question

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

2

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3

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5 : 6

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6

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Solution

The correct option is C $5 : 6$
$\begin{matrix} t a k e n = 2 f = 3 A B : w = 2 R (T_{o}) = 2 R T_{0} Δ u = \frac{f}{2} n R Δ T = \frac{3}{2} \times 2 \times r T_{o} = 3 R T_{0} B C : P T^{2} = c o n s t a n t p . {(\frac{p v}{π R})}^{2} = c o n s tan t \Rightarrow p^{3} v^{2} = c o n s tan t ∴ w = \frac{n R (T_{0} - 250)}{1 - \frac{2}{3}} = - 3 n R T_{0} Δ u = - 3 n R T_{0} Δ Q = - 6 n R T_{0} H e n c e, R a t i o : \frac{5}{6} H e n c e, t h e o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Two moles of an ideal mono-atomic gas is taken through a cyclic process as shown in the P−T diagram. In the process BC,PT2 = constant. Then the ratio of heat absorbed and heat released by the gas during the process AB and process BC respectively is:-

The correct option is C 5:6taken=2f=3AB:w=2R(To)=2RT0Δu=f2nRΔT=32×2×rTo=3RT0BC:PT2=constantp.(pvπR)2=constant⇒p3v2=constant∴w=nR(T0−250)1−23=−3nRT0Δu=−3nRT0ΔQ=−6nRT0Hence,Ratio:56Hence,theoptionCiscorrectanswer.

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