Physics Q23 – NEET-UG 2023

The temperature of a gas is

- 50^{\circ}

C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

Clue 1 of 3: Kinetic Theory of Gases

The root mean square (rms) speed of gas molecules is related to its absolute temperature.

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Ninja Strategy•Proportionality and Magnitude Estimation

Recognize that $v_{r m s} \propto \sqrt{T}$ . If speed increases by a factor of $X$ , temperature must increase by $X^{2}$ . An increase 'by 3 times' means the speed becomes 4 times the original, so the temperature must become $4^{2} = 16$ times the original absolute temperature. Quickly estimate which option is roughly 16 times the initial absolute temperature ( $223$ K).

🧠 Full Solution Path

Step 1: Convert Initial Temperature to Kelvin✦ Active

The initial temperature is given as $T_{1} = - 50^{\circ}$ C. To use the gas laws, temperature must be in Kelvin. Convert Celsius to Kelvin by adding $273$ :

T_{1} = - 50 + 273 = 223 K

Step 2: Relate RMS Speed to Absolute Temperature○ Expand

The root mean square (rms) speed of gas molecules is directly proportional to the square root of the absolute temperature ( $v_{r m s} \propto \sqrt{T}$ ). The problem states that the rms speed is "increased by 3 times". In physics, "increased by N times" means the new value is $(N + 1)$ times the original value. So, the final rms speed $v_{r m s 2}$ is $4$ times the initial rms speed $v_{r m s 1}$ .

\frac{v_{r m s 2}}{v_{r m s 1}} = 4

Using the proportionality, we can write:

\frac{v_{r m s 2}}{v_{r m s 1}} = \sqrt{\frac{T_{2}}{T_{1}}}

\text{Substituting the ratio of speeds:}

4 = \sqrt{\frac{T_{2}}{223}}

💡 Teacher's Secret Hint

Pay close attention to the phrasing "increased by 3 times" vs. "increased to 3 times". The former implies $X_{f i n a l} = (1 + 3) X_{i n i t i a l}$ .

Step 3: Calculate the Final Temperature○ Expand

Square both sides of the equation to solve for $T_{2}$ :

4^{2} = \frac{T_{2}}{223} ⟹ 16 = \frac{T_{2}}{223}

Now, calculate $T_{2}$ in Kelvin:

T_{2} = 16 \times 223 = 3568 K

Finally, convert $T_{2}$ back to Celsius, as the options are in Celsius:

T_{2} ({in}^{\circ} C) = 3568 - 273 = 3295^{\circ} C

This matches option (2).

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Physics Question 23 – NEET-UG 2023

Clue 1 of 3: Kinetic Theory of Gases

🧠 Full Solution Path