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BITSAT 2019
Exam
BITSAT
Year
2019
Subject
Physics
Chapter
Thermodynamics

BITSAT 2019 · Physics · Thermodynamics

The figure shows the graph of (pV)/(T) versus p for 2 × 10 ^ - 4 kg...

The figure shows the graph of pVT\dfrac{pV}{T} versus pp for 2×10−4kg2 \times {10}^{- 4}kg of hydrogen gas at two different temperatures, where p,Vp, V and TT represents pressure, volume and temperature respectively.Then, the value of pVT,\dfrac{pV}{T},& where the curve meet on the vertical axis, is

The figure shows the graph of (pV)/(T) versus p for 2 × 10 ^ - 4 kg of hydrogen gas at two different temperatures,

Options

  1. AR5\dfrac{R}{5}
  2. B2R3\dfrac{2R}{3}
  3. CR10\dfrac{R}{10}
  4. DR15\dfrac{R}{15}

Correct answer

C. R10\dfrac{R}{10}

Step-by-step solution

Number of molesmoles in 2×10−4kg2 \times {10}^{- 4}kg

μ=MassofhydrogeningramMolecularmass\mu = \dfrac{Massofhydrogeningram}{Molecularmass}

=2×10−4×1032= \dfrac{2 \times {10}^{- 4} \times {10}^{3}}{2}

=0.1mole= 0. 1mole

∴pV=\muRT∴ pV = \muRT

pVT=\muR=0.1×R=R10\dfrac{pV}{T} = \muR = 0. 1 \times R = \dfrac{R}{10}

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