Q29 / 90 · JEE Main 2013 (09 Apr Online) · Chemistry

By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from $5 \times 10^4 \mathrm{~cm} / \mathrm{s}$ to $10 \times 10^4 \mathrm{~cm} / \mathrm{s}$ ?

  • A
    Two
  • B
    Three
  • C
    Six
  • Correct answer
    D
    Four

Solution

r.m.s. velocity $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$ i.e., $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}$ $\frac{5 \times 10^4}{10 \times 10^4}=\frac{1}{2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}$ $\therefore \mathrm{T}_2=4 \mathrm{T}_1$

Hence, the correct answer is option: 'D'.

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