The temperature of a gas is raised from 27° C to 927° C. The root mean square speed

gets halved

gets doubled

is $(\sqrt{\frac{927}{27}})$ times the earlier value

remains the same

Solution

gets doubled

Given:-

T₁ = 27^oC = 300 K

T₂ = 927^oC = 1200 K

According to Kinetic theory of ideal gas

v_rms = ${(\bar{v^{2}})}^{\frac{1}{2}} = \sqrt{\frac{3 KT}{m}}$

v_rms = $\sqrt{(\frac{3 kT}{m})}$

Where m - mass of the molecules

$\bar{v^{2}}$ - mean of squared speed

⇒ v_rms = $\sqrt{T}$

∴ $\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

$\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{1200}}$

( v_rms)₂ = 2 ( v_rms )₁

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