1. 223 K 2. 669^° C 3. 3295^° C 4. 3097 K

Answer: 3. 3295^° C v_rms∝√(T) . The key reads \"increased by 3 times\" as the speed becoming v+3v=4v . T₁ = -50+273=223 K. v₂/v₁=4⇒ T₂=16T₁=16×223=3568 K. T₂ = 3568-273 = 3295^° C. The answer is 3295^° C. (If the speed simply became 3 times, T₂=9×223=2007 K, which is not an option.)

Topic: Kinetic interpretation of temperature, RMS speed· Kinetic Theory of Gases

The temperature of a gas is -50^° C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

1. 223 K

2. \(669^\circ\) C

3. \(3295^\circ\) C

4. 3097 K

Doubtly Asked 30 Oct 2026 NEET UG 2023 Practice test neet Save WhatsApp Report

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Answer: 3. \(3295^\circ\) C

\(v_{rms}\propto\sqrt{T}\). The key reads \"increased by 3 times\" as the speed becoming \(v+3v=4v\).

\(T_1 = -50+273=223\) K.
\(\dfrac{v_2}{v_1}=4\Rightarrow T_2=16T_1=16\times223=3568\) K.
\(T_2 = 3568-273 = 3295^\circ\)C.

The answer is \(3295^\circ\)C. (If the speed simply became 3 times, \(T_2=9\times223=2007\) K, which is not an option.)

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The temperature of a gas is -50^° C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

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