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Calculate the number of molecules in 2 x...

Calculate the number of molecules in `2 xx 10^(-6)m^(3)` of a perfect gas at `27^(@) C` and at a pressure of 0.01 mm of mercury. Mean KE of a molecules at `27^(@) C = 4xx10^(-11) J and g = 9.8 m//s^(2)`.

Text Solution

Verified by Experts

The correct Answer is:
8
Here V=`2xx10^(-6)m^3`
`P=h rho g =0.01 xx 13.6 xx10^3 xx9.8 =13.6 xx 98Nm^(-2)`
Total K.E. of the gas molecules,
`1/2Mv_(rms)^2=3/2 PV=3/2 xx 13.6 xx98xx2xx10^(-6)` J
`therefore` No. of molecules in the given volume
`=(Total K.E. of the gas molecules)/(K.E. per molecule)`
`(3xx13.6 xx 98xx10^(-6))/(4xx10^(-11))`
`=9996 xx10^(4)=10^8`
On comparing `10^8` with `10^A` then we get A=8
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