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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Thermodynamics

BITSAT 2019 · Physics · Thermodynamics

An ideal monoatomic gas at 300K expands adiabatically to twice its...

An ideal monoatomic gas at 300K300K expands adiabatically to twice its volume. The final temperature of gas is

Options

  1. A3002300\sqrt{2}
  2. B3003300\sqrt{3}
  3. C300(12)23\left(\dfrac{1}{2}\right)\dfrac{2}{3}
  4. D300(2)23\left(2\right)\dfrac{2}{3}

Correct answer

C. 300(12)23\left(\dfrac{1}{2}\right)\dfrac{2}{3}

Step-by-step solution

For an ideal monatomic gas, the ratio of specific heat capacity at constant pressure to specific heat capacity at constant volume is

γ=53\gamma = \dfrac{5}{3}

In an adiabatic process,

TVγ−1=TV^{\gamma - 1} = constant

i.e. TfVfγ−1=TiViγ−1⇒Tf=Ti(ViVf)γ−1=300(12)53−1{T}_{f}{V}_f^{\gamma - 1} = {T}_{i}{V}_i^{\gamma - 1} ⇒ {T}_{f} = {T}_{i}\left(\dfrac{Vi}{Vf}\right)\gamma - 1 = 300\left(\dfrac{1}{2}\right)\dfrac{5}{3} - 1

=300(12)2/3= 300\left(\dfrac{1}{2}\right)^{2} / 3

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