The temperature of a gas is –68°C. To what temperature should it be heated, so that the r.m.s. velocity of the molecules be doubled?

  • Option 1)

    357°C

  • Option 2)

    457°C

  • Option 3)

    547°C

  • Option 4)

    820°C

 

Answers (1)

As we learnt in 

Root mean square velocity -

V_{rms}= \sqrt{\frac{3RT}{M}}

= \sqrt{\frac{3P}{\rho }}
 

- wherein

R = Universal gas constant

M = molar mass

P = pressure due to gas

\rho = density

 

 V_{rms}\ \propto \ \sqrt{T}

\therefore\ \frac{V_{2}}{V_{1}}= \sqrt{\frac{T_{2}}{T_{1}}}

\Rightarrow \frac{T_{2}}{T_{1}}= \left ( \frac{V_{2}}{V_{1}} \right )^{2}

\Rightarrow T_{2} = \left ( \frac{2V_{1}}{V_{1}} \right )^{2}\ T_{1} = 4 \ T_{1}= 4\left ( 273-68 \right )

T_{2}= 820\ K= 547^{\circ}\ C


Option 1)

357°C

This option is incorrect

Option 2)

457°C

This option is incorrect

Option 3)

547°C

This option is correct

Option 4)

820°C

This option is incorrect

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions