What is the pressure exerted by 32.00 g of Oxygen gas in a 20. L container at 30.00 C? Use R = 0.0821 L atm/mol k?

3 Answers
May 20, 2017

Answer:

#1.24# #"atm"#

Explanation:

First, let's determine the number of mole of oxygen gas.

Using #n = frac(m)(M)#:

#Rightarrow n("O") = (frac(32.00)(2 times 15.99))# #"mol"#

#Rightarrow n("O") = (frac(32.00)(31.98))# #"mol"#

#therefore n("O") = 1.00# #"mol"#

Then, let's convert the units of the given temperature to #"K"#:

#Rightarrow T_((""^(@)"" "C")) = T_(("K")) - 273.15#

#Rightarrow 30.00 = T_(("K")) - 273.15#

#Rightarrow T_(("K")) = 303.15#

#therefore 30.00# #""^(@)"" "C"# #= 303.15# #"K"#

Now, let's substitute all relevant values into the equation #P V = n R T#:

#Rightarrow P times 20.0# #"L"# #= 1.00# #"mol"# #times 0.0821# #"L atm mol"^(- 1) "K"^(- 1)# #times 303.15# #"K"#

#Rightarrow P times 20.0 = 1.00 times 0.0821# #"atm"# #times 303.15#

#Rightarrow P = (frac(0.0821 times 303.15)(20.0))# #"atm"#

#therefore P = 1.24# #"atm"#

Therefore, the pressure exerted by the oxygen gas is #1.24# #"atm"#.

May 20, 2017

Answer:

#1.24# atm, rounded to two significant figures

Explanation:

Note: mols and moles are the same things; formatting issues prevent me from using the same word throughout the answer
We will use the ideal gas equation, which is as follows:

#PV = nRT#

#P# is pressure (atm)
#V# is volume( liters)
#n# is moles of substance (#(Liters*atm)/(mols*K)#)
#R# is the constant #0.0821#
#T# is temperature (kelvin).

We are solving for #P# in this equation, so the formula is reworked as so:
#P = (nRT)/V#
Now, we find the rest of the numbers to plug in

#n#: We convert Grams of #O_"2"# to moles using dimensional analysis: #32.00g O_2 * (1 mol O_2)/(32.00g O_2) = 1 mol O_2#

#R#: We use the constant #0.0821#

#T#: We convert the temperature from Celcius to Fahrenheit, using the formula #K = ^oC + 273#
#K = 30 + 273 = 303K#

#V#: We use the given liters, which is #20L#

Now, we solve.

First by plugging in numbers:

#P = (nRT)/V#
=> #P = ((1 mol)(0.0821(Liters*atm)/(mols*K))(303K))/(20L)#

Which is the same as:

#P = ((1 mol)(0.0821(Liters*atm))(303K))/(20L*mols*K)#

Liters, moles, and Kelvin cancel out, giving us:

#P = ((1)(0.0821 atm)(303))/(20)#

Now we simplify to get our answer:

#P = ((1)(0.0821 atm)(303))/(20)#

=> #P = (24.88atm)/(20)#

=> #P = 1.24 atm#

May 20, 2017

Answer:

The pressure is 1.2 atm.

Explanation:

This is an ideal gas law problem. The equation is:

#PV=nRT#,

where #P# is pressure, #V# is volume, #n# is moles, #R# is a gas constant, and #T# is temperature in Kelvins.

You have been given the mass of #"O"_2#, but the equation requires moles. Determine the mol #"O"_2# by multiplying its given mass by the inverse of its molar mass (31.998 g/mol).

#32.00color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="1.000 mol O"_2"#

You have been given temperature in degrees Celsius, but gas problems require the temperature to be in Kelvins. Convert #30.00^@"C"# to Kelvins by adding #273.15#.

#30.00^@"C" + 273.15="303.15 K"#

#color(blue)("Now organize your data."#

Given/Known

#V="20. L"=2.0xx10^2color(white)(.)"L"#

#n="1.000 mol"#

#R="0.0821 L atm K"^(-1) "mol"^(-1)"#

#T="303.15 K"#

#color(blue)("Solution."#
Rearrange the equation to isolate #P#. Insert your data and solve.

#P=(nRT)/V#

#P=(1.000color(red)cancel(color(black)("mol"))xx0.0821color(red)cancel(color(black)("L")) "atm" color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx303.15color(red)cancel(color(black)("K")))/(2.0xx10^2color(red)cancel(color(black)("L")))="1.2 atm"# rounded to two significant figures)