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NEET
Exam
NEET
Subject
Physics
Chapter
Thermodynamics

NEET · Physics · Thermodynamics

The pressure versus temperature graph of an ideal gas is shown in...

The pressure versus temperature graph of an ideal gas is shown in the figure below. If the density of gas at point AA is ρ0{ρ}_{0}, then the density of the gas at point BB will be

The pressure versus temperature graph of an ideal gas is shown in the figure below. If the density of gas at point A is

Options

  1. A2ρ02{ρ}_{0}
  2. B23ρ0\dfrac{2}{3}{ρ}_{0}
  3. C32ρ0\dfrac{3}{2}{ρ}_{0}
  4. D3ρ03{ρ}_{0}

Correct answer

B. 23ρ0\dfrac{2}{3}{ρ}_{0}

Step-by-step solution

Given, density of gas at point A,ρA=ρ0A, {ρ}_{A} = {ρ}_{0}

General equation for an ideal gas

pV=nRTpV = nRT

⇒pV=mMRT⇒ pV = \dfrac{m}{M}RT [m→m \tomass, M→M \to molecular mass]

⇒p=mV⋅RM⋅T⇒ p = \dfrac{m}{V} \cdot \dfrac{R}{M} \cdot T

⇒p=ρ⇒ p = ρ\left(\dfrac{R}{M}\right)T . . . (i)\left(i\right) [where density ρ=mVρ = \dfrac{m}{V}]

Applying Equation. (i) at point AA and BB, respectively.

p=ρA(RM)T0...(ii)p = {ρ}_{A}\left(\dfrac{R}{M}\right){T}_{0} . . . \left(ii\right)

2p=ρB(RM)3T0...(iii)2p = {ρ}_{B}\left(\dfrac{R}{M}\right)3{T}_{0} . . . \left(iii\right)

From Equations. (ii) and (iii), we get

12=ρAρB⋅13\dfrac{1}{2} = \dfrac{{ρ}_{A}}{{ρ}_{B}} \cdot \dfrac{1}{3}

⇒ρB=23ρA=23ρ0⇒ {ρ}_{B} = \dfrac{2}{3}{ρ}_{A} = \dfrac{2}{3}{ρ}_{0}

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