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Heat and Thermodynamics question

2024 · 29 Jan · Shift 1 · Q75
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  5. /2024 · 29 Jan · Shift 1 · Q75

Heat and Thermodynamics question

2024 · 29 Jan · Shift 1 · Q75

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Two vessels AAA and BBB are of the same size and are at same temperature. A contains 1 g1 \mathrm{~g}1 g of hydrogen and BBB contains 1 g1 \mathrm{~g}1 g of oxygen. PA\mathrm{P}_{\mathrm{A}}PA​ and PB\mathrm{P}_{\mathrm{B}}PB​ are the pressures of the gases in A\mathrm{A}A and B\mathrm{B}B respectively, then PAPB\frac{P_A}{P_B}PB​PA​​ is:
  1. A
    4
  2. B
    32
  3. C
    8
  4. D
    16
View written solutionFree

Correct answer: D

  1. Use the ideal gas law

For a gas in a vessel, PV=nRTPV=nRTPV=nRT

Since vessels AAA and BBB have:

  • same volume VVV
  • same temperature TTT

we get P∝nP \propto nP∝n

So, PAPB=nAnB\frac{P_A}{P_B}=\frac{n_A}{n_B}PB​PA​​=nB​nA​​

  1. Find moles in each vessel

Vessel AAA: Hydrogen

Hydrogen gas is H2\mathrm{H_2}H2​, so its molar mass is MH2=2 g/molM_{H_2}=2\ \text{g/mol}MH2​​=2 g/mol

Given mass = 1 g1\,\text{g}1g, nA=12 moln_A=\frac{1}{2}\,\text{mol}nA​=21​mol

Vessel BBB: Oxygen

Oxygen gas is O2\mathrm{O_2}O2​, so its molar mass is MO2=32 g/molM_{O_2}=32\ \text{g/mol}MO2​​=32 g/mol

Given mass = 1 g1\,\text{g}1g, nB=132 moln_B=\frac{1}{32}\,\text{mol}nB​=321​mol

  1. Calculate pressure ratio

PAPB=nAnB=12132=12×32=16\frac{P_A}{P_B}=\frac{n_A}{n_B}=\frac{\frac{1}{2}}{\frac{1}{32}}=\frac{1}{2}\times 32=16PB​PA​​=nB​nA​​=321​21​​=21​×32=16

  1. Match with options

The correct option is: 16\boxed{16}16​

So, Option D is correct.

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