JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Heat & Thermodynamics

The temperature of a gas having

2.0 \times 1 0^{25}

molecules per cubic meter at 1.38 atm (Given,

k = 1.38 \times 1 0^{- 23}

^{- 1}

) is :

(A)500 K✓ Correct
(B)200 K
(C)100 K
(D)300 K

Answer: (A)

Solution

The ideal gas law

P V = n R T

, written using number density

n_{v} = N / V

and Boltzmann's constant

k

, becomes

P = n_{v} k T

Substituting

P = 1.38 atm = 1.38 \times 1.01 \times 1 0^{5} Pa

n_{v} = 2.0 \times 1 0^{25} m^{- 3}

k = 1.38 \times 1 0^{- 23} JK^{- 1}

1.38 \times 1.01 \times 1 0^{5} = 2.0 \times 1 0^{25} \times 1.38 \times 1 0^{- 23} \times T

1.01 \times 1 0^{5} = 2 \times 1 0^{2} \times T

T = \frac{1.01 \times 1 0 ^{3}}{2} \approx 500 K

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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