Gaseous Mixture and Volume! Anyone know how to get this problem started!?

Question

What volume will a mixture of 0.42 mole H2 and 0.84 mole Ar occupy at 650.0 torr and 16.0 °C?

Answer 1

Use the ideal gas law PV = nRT

You don't have to solve seperately for H2 and Ar, based on the ideal gas law, you can add the moles of gas together.

Convert the pressure from torr to atm (760 torr = 1 atm.) then you can use 0.0821 for the "R" constant. Don't forget to convert the temperature to Kelvin (add 273)

Answer 2

Ideal Gas Law PV=nRT, to obtain V you have to manipulate the equation, therefore V=nRT/P.
Given:
P=650 torr, this uses atm so convert torr into atm,
1 atm=760 torr; 650 torr * 1 atm/760 torr, cancel 650 torr with 760 torr
P=0.855 atm
n=mole H2+mole Ar
R=[universal gas constant]0.08206 L*atm/mol*K
T=16.0C, this uses K so convert C into K, add 273.15

You have everything, mere substitution of the values I have provided.

TC! GOD BLESS!

Answer 3

Apply PV=nRT were n= Nº Moles H2 +Nª moles Ar

T = 16 + 273
R= 0.082
P= 655/760