Your idea matters-12

Classical Mechanics Level 2

If you compress a diatomic noble gas by adiabatic process in STP and make the volume half than the previous,then what will be the final pressure in units of meters of mercury ?


The answer is 2.0057.

Fahim Muhtamim by Fahim Muhtamim , 17, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Steven Chase
Oct 12, 2019

For an adiabatic process,

P 1 V 1 γ = P 2 V 2 γ P 2 = P 1 ( V 1 V 2 ) γ P_1 V_1^\gamma = P_2 V_2^\gamma \\ P_2 = P_1 \Big(\frac{V_1}{V_2} \Big)^\gamma

For a diatomic gas, γ = 7 / 5 \gamma = 7/5 . We also know that the volume gets cut in half. Therefore (assuming a "meters of mercury" convention):

P 2 = ( 0.76 ) 2 7 / 5 = 2.0057 P_2 = (0.76) \, 2^{7/5} = 2.0057

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...