Question

The rms velocity of gas molecules of a given amount of a gas at 27°C and 1.0*10^5 N/m² pressure is 200 m/sec. Of temperature and pressure are respectively 127°Cand 0.5*10^5N/m², the rms velocity will be

Answer 1

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Answer 2

Answer:

The rms velocity will be 230.94 m/s.

Explanation:

Given that,

rms velocity $v_{1}= 200\ m/s$

Pressure $P = 1.0\times10^{5}\ N/m^2$

Temperature $T=27^{\circ}C$

$T_{2}=127^{\circ}C$

We know that,

The rms velocity of gas molecule is directly proportional to the square root of temperature.

$v_{rms}\propto\sqrt{T}$

For two cases,

$\dfrac{v_{rms}_{1}}{v_{rms}_{2}}=\sqrt{\dfrac{T_{1}}{T_{2}}}$

$\dfrac{200}{v_{rms}_{2}}=\sqrt{\dfrac{27+273}{127+273}}$

$v_{rms}_{2}=\dfrac{200\times\sqrt{4}}{\sqrt{3}}$

$v_{rms}_{2}=\dfrac{400}{\sqrt{3}}\ m/s$

$v_{rms}_{2}=230.94\ m/s$

Hence, The rms velocity will be 230.94 m/s.