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Heat and Thermodynamics question

2024 · 29 Jan · Shift 2 · Q70
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  5. /2024 · 29 Jan · Shift 2 · Q70

Heat and Thermodynamics question

2024 · 29 Jan · Shift 2 · Q70

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The temperature of a gas having 2.0×10252.0 \times 10^{25}2.0×1025 molecules per cubic meter at 1.38 atm1.38 \mathrm{~atm}1.38 atm(Given, k=1.38×10−23JK−1\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1}k=1.38×10−23JK−1) is :
  1. A
    500 K
  2. B
    300 K
  3. C
    200 K
  4. D
    100 K
View written solutionFree

Correct answer: A

  1. Use the ideal gas relation in molecular form

For an ideal gas,

P=nkTP = n k TP=nkT

where:

  • PPP = pressure
  • nnn = number of molecules per unit volume
  • kkk = Boltzmann constant
  • TTT = temperature

So,

T=PnkT = \frac{P}{nk}T=nkP​

  1. Given data

n=2.0×1025 m−3n = 2.0 \times 10^{25}\ \text{m}^{-3}n=2.0×1025 m−3 P=1.38 atmP = 1.38\ \text{atm}P=1.38 atm k=1.38×10−23 J K−1k = 1.38 \times 10^{-23}\ \text{J K}^{-1}k=1.38×10−23 J K−1

Convert pressure into SI unit:

1 atm≈1.013×105 Pa1\ \text{atm} \approx 1.013 \times 10^5\ \text{Pa}1 atm≈1.013×105 Pa

Hence,

P=1.38×1.013×105P = 1.38 \times 1.013 \times 10^5P=1.38×1.013×105 P≈1.398×105 PaP \approx 1.398 \times 10^5\ \text{Pa}P≈1.398×105 Pa

  1. Substitute into the formula

T=1.398×105(2.0×1025)(1.38×10−23)T = \frac{1.398 \times 10^5}{(2.0 \times 10^{25})(1.38 \times 10^{-23})}T=(2.0×1025)(1.38×10−23)1.398×105​

First calculate the denominator:

(2.0×1025)(1.38×10−23)=2.76×102=276(2.0 \times 10^{25})(1.38 \times 10^{-23}) = 2.76 \times 10^2 = 276(2.0×1025)(1.38×10−23)=2.76×102=276

Now,

T=1.398×105276T = \frac{1.398 \times 10^5}{276}T=2761.398×105​

T≈506.5 KT \approx 506.5\ \text{K}T≈506.5 K

  1. Choose the nearest option

T≈500 KT \approx 500\ \text{K}T≈500 K

So the correct option is A.

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