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Heat and Thermodynamics question

2024 · 9 Apr · Shift 1 · Q65
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  5. /2024 · 9 Apr · Shift 1 · Q65

Heat and Thermodynamics question

2024 · 9 Apr · Shift 1 · Q65

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The volume of an ideal gas (γ=1.5)(\gamma=1.5)(γ=1.5) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is :
  1. A
    45\frac{4}{5}54​
  2. B
    855\frac{8}{5 \sqrt{5}}55​8​
  3. C
    25\frac{2}{\sqrt{5}}5​2​
  4. D
    1625\frac{16}{25}2516​
View written solutionFree

Correct answer: B

  1. Use the adiabatic relation

For an ideal gas undergoing an adiabatic process, PVγ=constantPV^\gamma = \text{constant}PVγ=constant

So, PiViγ=PfVfγP_i V_i^\gamma = P_f V_f^\gammaPi​Viγ​=Pf​Vfγ​

We need PiPf=VfγViγ=(VfVi)γ\frac{P_i}{P_f} = \frac{V_f^\gamma}{V_i^\gamma} = \left(\frac{V_f}{V_i}\right)^\gammaPf​Pi​​=Viγ​Vfγ​​=(Vi​Vf​​)γ

  1. Substitute the given values

Given:

  • γ=1.5=32\gamma = 1.5 = \frac{3}{2}γ=1.5=23​
  • Vi=5V_i = 5Vi​=5 L
  • Vf=4V_f = 4Vf​=4 L

Hence, PiPf=(45)3/2\frac{P_i}{P_f} = \left(\frac{4}{5}\right)^{3/2}Pf​Pi​​=(54​)3/2

  1. Simplify

(45)3/2=(45)3=(25)3\left(\frac{4}{5}\right)^{3/2} = \left(\sqrt{\frac{4}{5}}\right)^3 = \left(\frac{2}{\sqrt{5}}\right)^3(54​)3/2=(54​​)3=(5​2​)3

Now, (25)3=855\left(\frac{2}{\sqrt{5}}\right)^3 = \frac{8}{5\sqrt{5}}(5​2​)3=55​8​

  1. Match with the options

Thus, PiPf=855\frac{P_i}{P_f} = \frac{8}{5\sqrt{5}}Pf​Pi​​=55​8​

This matches Option B.

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