If the temperature of a gas is increased from 27- C to 927-C- the root mean square speed of its molecules

The root mean square speed is given by V_rms=√(3kT/m) i.e. V_rms∝ T. Initial temperature T_1=27+273=300K. Final temperature T_2=927+273=1200K. Since temperature ...

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Standard XII

Physics

Root Mean Square Speed

If the temper...

Question

If the temperature of a gas is increased from $27^{\circ}$ C to $927^{\circ}$ C, the root mean square speed of its molecules

becomes half

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is doubled

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becomes 4 times

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remains unchanged

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Solution

The correct option is B
is doubled

The root mean square speed is given by
$V_{r m s} = \sqrt{\frac{3 k T}{m}}$
i.e. $V_{r m s} \propto T$ . Initial temperature $T_{1} = 27 + 273 = 300 K .$ Final temperature $T_{2} = 927 + 273 = 1200 K .$ Since temperature is increased by 4 times, the speed is doubled. Hence the correct choice is (b)

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If the temperature of a gas is increased from 27∘ C to 927∘C, the root mean square speed of its molecules

The correct option is B is doubled The root mean square speed is given by Vrms=√3kTm i.e. Vrms∝T. Initial temperature T1=27+273=300K. Final temperature T2=927+273=1200K. Since temperature is increased by 4 times, the speed is doubled. Hence the correct choice is (b)

If the temperature of a gas is increased from $27^{\circ}$ C to $927^{\circ}$ C, the root mean square speed of its molecules