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Heat and Thermodynamics question

2021 · 25 Feb · Shift 2 · Q58
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Heat and Thermodynamics question

2021 · 25 Feb · Shift 2 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V2 = 2V1 then the ratio of temperature T2/T1 is : JEE Main 2021 (Online) 25th February Evening Shift Physics - Heat and Thermodynamics Question 267 English
  1. A
    2\sqrt 22​
  2. B
    12{1 \over {\sqrt 2 }}2​1​
  3. C
    12{1 \over 2}21​
  4. D
    2
View written solutionFree

Correct answer: A

  1. Use the ideal gas law for one mole: PV=RTPV = RTPV=RT Hence, T∝PVT \propto PVT∝PV for fixed number of moles.

  2. Read the process from the given PPP-VVV curve.
    The curve shown is a rectangular hyperbola-like relation corresponding to P∝1VP \propto \frac{1}{\sqrt V}P∝V​1​ so that PV=constantP\sqrt V = \text{constant}PV​=constant

  3. Therefore, between states 1 and 2: P1V1=P2V2P_1\sqrt{V_1} = P_2\sqrt{V_2}P1​V1​​=P2​V2​​ which gives P2P1=V1V2\frac{P_2}{P_1} = \sqrt{\frac{V_1}{V_2}}P1​P2​​=V2​V1​​​

  4. Now, T2T1=P2V2P1V1=P2P1⋅V2V1\frac{T_2}{T_1} = \frac{P_2V_2}{P_1V_1} = \frac{P_2}{P_1}\cdot \frac{V_2}{V_1}T1​T2​​=P1​V1​P2​V2​​=P1​P2​​⋅V1​V2​​ Substitute the pressure ratio: T2T1=V1V2⋅V2V1=V2V1\frac{T_2}{T_1} = \sqrt{\frac{V_1}{V_2}}\cdot \frac{V_2}{V_1} = \sqrt{\frac{V_2}{V_1}}T1​T2​​=V2​V1​​​⋅V1​V2​​=V1​V2​​​

  5. Given: V2=2V1V_2 = 2V_1V2​=2V1​ so, T2T1=2\frac{T_2}{T_1} = \sqrt{2}T1​T2​​=2​

  6. Check options:

    • A: 2\sqrt 22​ ✅
    • B: 12\dfrac{1}{\sqrt 2}2​1​
    • C: 12\dfrac{1}{2}21​
    • D: 222

So the correct answer is A.

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