A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm, at a fixed temperature.

A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 m

1.	A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm, at a fixed temperature.
Answer» Solution :`P_(NE)=X_(Ne)^(-)P_("Total")` `X_(Ne)=(n_(Ne))/(n_(Ne)+n_(Ar)+n_(XE))=(4.76)/(4.76+0.74+2.5)=0.595` `X_(Ar)=(n_(Ar))/(n_(Ne)+n_(Ar)+n_(Xe))=(0.74)/(4.76+0.74+2.5)=0.093` `X_(Xe)=(n_(Xe))/(n_(Ne)+n_(Ar)+n_(Xe))=(2.5)/(4.76+0.74+2.5)=0.312` `P_(Ne)=X_(Ne)P_("Total")=0.595xx2=1.19` ATM. `P_(Ar)=X_(Ar)P_(Total)=0.093xx2=0.186atm` `P_(Xe)=X_(Xe)P_(Total)=0.312xx2=0.624atm`.