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The temperature of a gas having 2.0 × 10²⁵ molecules per cubic meter at 1.38 atm is (Given, k = 1.38 × 10⁻²³ J K⁻¹)

Asked in JEE Main 29th Jan 2nd Shift 2024 · Number of molecules and number density

Answer: (3) 500 K

Step-by-step solution

For an ideal gas written per molecule, P=nk_BT with n the number density.

T=P/(nk_B).

Taking the stated pressure as 1.38×10⁵ Pa and n=2.0×10²⁵ m⁻³:

nk_B=2.0×10²⁵×1.38×10⁻²³=276.

T=(1.38×10⁵)/(276)=500 K.

Why the other options are wrong

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