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» <html><body><p></p>Solution :`P_(<a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a>)=X_(Ne)^(-)P_("Total")` <br/> `X_(Ne)=(n_(Ne))/(n_(Ne)+n_(Ar)+n_(<a href="https://interviewquestions.tuteehub.com/tag/xe-747964" style="font-weight:bold;" target="_blank" title="Click to know more about XE">XE</a>))=(4.76)/(4.76+0.74+2.5)=0.595` <br/> `X_(Ar)=(n_(Ar))/(n_(Ne)+n_(Ar)+n_(Xe))=(0.74)/(4.76+0.74+2.5)=0.093` <br/> `X_(Xe)=(n_(Xe))/(n_(Ne)+n_(Ar)+n_(Xe))=(2.5)/(4.76+0.74+2.5)=0.312` <br/> `P_(Ne)=X_(Ne)P_("Total")=0.595xx2=1.19` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>. <br/> `P_(Ar)=X_(Ar)P_(Total)=0.093xx2=0.186atm` <br/> `P_(Xe)=X_(Xe)P_(Total)=0.312xx2=0.624atm`.</body></html>