We mix 45 grams of oxygen gas with 183 grams of argon gas in a volume of 1697mL at 160C. What will be the fin
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We mix 45 grams of oxygen gas with 183 grams of argon gas in a volume of 1697mL at 160C. What will be the fin

[From: ] [author: ] [Date: 12-04-17] [Hit: ]
We shall now use the ideal gas equation PV= nRT where P is the total pressure exerted by the gas particles, V is the volume of the container, n is the total number of moles of gas particles, R is the molar gas constant(8.31J/mol/K) and T is the temperature.P= nRT/V = (5.......
We mix 45 grams of oxygen gas with
183 grams of argon gas in a volume of 1697mL
at 160C. What will be the final pressure of
the gas mixture?
Answer in units of atm

-
Okay, first of all, let's convert all units to SI units.

1.000mL= 1.000x10⁻⁶m³
1697mL= (1.000x10⁻⁶)(1697)
= 1.697x10⁻³m³
160°C= (160+273)K
= 433K

Relative Molecular Mass(RMM) of O₂
= 2(RAM of O)
= 2(16)
= 32

32g= 1 mole of O₂
45g= 45/32
= 1.40625 moles of O₂

Relative Atomic Mass(RAM) of Ar
= 39.95

39.95g= 1 mole of Ar
183g= 183/39.95
= 4.580725907 moles of Ar

Total number of moles of O₂ and Ar
= 5.986975907

We shall now use the ideal gas equation PV= nRT where P is the total pressure exerted by the gas particles, V is the volume of the container, n is the total number of moles of gas particles, R is the molar gas constant(8.31J/mol/K) and T is the temperature.

P= nRT/V
= (5.986975907)(8.31)(433)/1.697x10⁻³
= 1.269447043x10⁷N/m²

1.01x10⁵N/m²= 1.00atm
1.269447043x10⁷N/m²= 1.269447043x10⁷/1.01x10⁵
= 125.687826atm
= 126atm correct to 3sf

Thus, the final pressure will be 126atm. I hope this helps and feel free to send me an e-mail if you have any doubts!
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