Question

When the temperature of an ideal gas is increased by 600 K, the velocity of sound in the gas becomes $ \sqrt{3} $ times the initial velocity in it. The initial temperature of the gas is:

• A. $ -73{}^\circ C $
• B. $ 27{}^\circ C $
• C. $ 127{}^\circ C $
• D. $ 327{}^\circ C $

Answer 1

Answer: (B)

Let initial temperature is T ie $ {{T}_{1}}=T $ Then, $ {{T}_{2}}=T+600 $ Velocity of sound in gas $ v\propto \sqrt{T} $ $ \frac{{{v}_{2}}}{{{v}_{1}}}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}} $ According to questions $ {{v}_{2}}=\sqrt{3}{{v}_{1}} $ $ \frac{\sqrt{3}{{v}_{1}}}{{{v}_{1}}}=\sqrt{\frac{T+600}{T}} $ $ \sqrt{3}=\sqrt{\frac{T+600}{T}} $ $ \therefore $ $ 3T=T+600 $ $ T=300\,K $ $ =300-273\,=27{{\,}^{o}}C $