Calculate the pressure (in atm) needed to compress 500cm^3 oxygen gas at 750mmhg to 100cm^3

Pressure in terms of Limited

Explanation:

4

We then have









 







1.6.

Dalton’s law

This law states that the total pressure of a mixture of gases occupying a given volume isequal to the sum of the partial pressures of the individual gases in the mixture. Thepartial pressure of gas in a mixture is the pressure the gas would exert if it occupied thecontainer alone. It is important to mention

that Dalton’s law of partial pressures applies

to gases which do not react chemically, i.e. a mixture of gases. For example, a mixture ofhydrogen and oxygen gases obeys this law if they have not reacted to form acompound, e.g. water or water vapour.

Mathematically, Dalton’s law can be written as

:



















where

P

T

is the total pressure of the mixture and





,





,





, … are the pressures of the

gases



,



,



, … in the mixture.

2.

The Ideal Gas Equation

This equation is a combination of the above laws

(excluding Dalton’s law)

whichestablishes relationship between temperature, pressure, volume and number ofmolecules in a given sample of gas.

Combining Boyle’s, Charles’ and Avogadro’s

lawswill produce the ideal gas equation as follows

  





    

    

By multiplying the right-hand sides of the above expressions, we have

()

  

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml