Question

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?

• A. 223 K
• B. 669°C
• C. 3295°C
• D. 3097 K

Answer 1

The Correct Option is C

We know that,
\(V_{rms}\propto \sqrt{T}\)

\(\frac{V_{1}}{V_{2}}=\sqrt{\frac{T_{1}}{T_{2}}}\)
\(⇒\) let the initial speed be \(v\).
As the speed increases by 3 times,
So, the final speed \(= 4v\)
\(⇒ \frac{v}{4v}\)

\(= \sqrt{\frac{223}{T}}\)
\(T=3568\text{ K}\)
So, the temperature in \(\degree{C} = 3568 - 273 = 3295\degree{C}\)

The correct answer is (C): \(3295°C\).

Answer 2

The new velocity (V’_rms) has been increased by 3 times the old velocity (V_rms).
Therefore, \(V’_{rms} = V_{rms} + 3 V_{rms} = 4V_{rms}\)
And, we also know V_rms is directly proportional to \(\sqrt{T}\)
Given, the initial temperature (T) = -50°C
Converting the temperature (T) to Kelvin: -50°C + 273 = 223K
Therefore initial temperature (T) in kelvin = 223K
If the speed increases to 4 times, the temperature should also increase by 16 times.
\(\frac{V’_{rms}}{V_{rms}} = \sqrt{\frac{T'}{T}}\)

\(\frac{4V_{rms}}{V_{rms}} = \sqrt \frac{T'}{223}\)
Therefore, Final Temperature (T’) \(= 16 \times 223 = 3568 K\)
Hence, Final Temperature (T’) in \(\degree C = 3568-273 = 3295 \degree C\)

So, the correct option is (C): \(3295 \degree C\)