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Heat and Thermodynamics question

2021 · 26 Aug · Shift 2 · Q49
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  5. /2021 · 26 Aug · Shift 2 · Q49

Heat and Thermodynamics question

2021 · 26 Aug · Shift 2 · Q49

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A cylindrical container of volume 4.0 ×\times× 10 −-− 3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is : [Take gas constant as 8.3 J mol −-− 1 K −-− 1]
  1. A
    249 ×\times× 101 Pa
  2. B
    24.9 ×\times× 103 Pa
  3. C
    24.9 ×\times× 105 Pa
  4. D
    24.9 Pa
View written solutionFree

Correct answer: C

  1. Use the ideal gas law for a mixture

For a mixture of ideal gases, PV=nRTPV = nRTPV=nRT where nnn is the total number of moles.

  1. Calculate total moles

Given:

  • Hydrogen =1= 1=1 mole
  • Carbon dioxide =2= 2=2 moles

So, n=1+2=3n = 1 + 2 = 3n=1+2=3

  1. Substitute the given values

Given:

  • V=4.0×10−3 m3V = 4.0 \times 10^{-3}\ \text{m}^3V=4.0×10−3 m3
  • T=400 KT = 400\ \text{K}T=400 K
  • R=8.3 J mol−1K−1R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}R=8.3 J mol−1K−1

Using P=nRTVP = \frac{nRT}{V}P=VnRT​

P=3×8.3×4004.0×10−3P = \frac{3 \times 8.3 \times 400}{4.0 \times 10^{-3}}P=4.0×10−33×8.3×400​

  1. Compute numerator

3×8.3×400=99603 \times 8.3 \times 400 = 99603×8.3×400=9960

So, P=99604.0×10−3P = \frac{9960}{4.0 \times 10^{-3}}P=4.0×10−39960​

  1. Calculate pressure

P=9960×14.0×10−3P = 9960 \times \frac{1}{4.0 \times 10^{-3}}P=9960×4.0×10−31​

14.0×10−3=250\frac{1}{4.0 \times 10^{-3}} = 2504.0×10−31​=250

Hence, P=9960×250=2.49×106 PaP = 9960 \times 250 = 2.49 \times 10^6\ \text{Pa}P=9960×250=2.49×106 Pa

  1. Match with options

2.49×106 Pa=24.9×105 Pa2.49 \times 10^6\ \text{Pa} = 24.9 \times 10^5\ \text{Pa}2.49×106 Pa=24.9×105 Pa

This matches Option C.

Final Answer

Option C: 24.9×105 Pa24.9 \times 10^5\ \text{Pa}24.9×105 Pa

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