What is the pressure in #mm*Hg# if a #4.0*g# mass of oxygen gas at #303*K# was confined to a container whose volume of #3000*mL#?

Question

Henry Antrim · Accepted Answer

#P=1.04*atm.......#

Using the Ideal Gas Equation, we can obtain.....

#P=(nRT)/V=((4.00*g)/(32.00*g*mol^-1)xx0.0821*(L*atm)/(K*mol)xx303*K)/(3000*mLxx10^-3*L*mL^-1)#

#=1.04*atm#

Now had the pressure been UNDER #1*atm# I could use the relationship #1*atm-=760*mm*Hg#; i.e. an atmosphere will support a column of mercury that is so high. You do not measure a pressure that is greater than #1*atm# in #mm*Hg#. Whoever set this problem was ignorant.........

Wyatt Adkins · Answer

undefined To find the pressure, we can use the ideal gas law equation: PV = nRT.
First, convert grams to moles: $ n = \frac{m}{M} $.
$ n = \frac{4.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.125 \, \text{mol} $.
Next, plug the values into the ideal gas law equation: $ P = \frac{nRT}{V} $.
$ P = \frac{(0.125 \, \text{mol})(0.08206 \, \text{atm} \cdot \text{L/mol} \cdot \text{K})(303 \, \text{K})}{3.00 \, \text{L}} $.
$ P = 3.74 \, \text{atm} $.
To convert atm to mmHg, use the conversion factor: 1 atm = 760 mmHg.
$ P = 3.74 \, \text{atm} \times 760 \, \text{mmHg/atm} $.
$ P = 2842.4 \, \text{mmHg} $.