A #0.275*L# volume of an unspecified gas exerts a pressure of #732.6*mm*Hg# at a temperature of #-28.2# #""^@C#. What is the molar quantity of this gas?

Question

A #0.275*L# volume of an unspecified gas exerts a pressure of #732.6*mm*Hg# at a temperature of #-28.2# #""^@C#.  What is the molar quantity of this gas?

Cameron Andrews · Accepted Answer

Approx. #0.013*mol# We need to know that #1*atm# will support a column of mercury #760*mm# high, and thus we can use the length of a mercury column to express pressure. And thus #P=(732.6*mm*Hg)/(760*mm*Hg*atm^-1)=??*atm# And we use the #"Ideal Gas Law"#, #n=(PV)/(RT)# #n=(PV)/(RT)=((732.6*mm*Hg)/(760*mm*Hg*atm^-1)xx0.275*L)/(0.0821*L*atm*K^-1*mol^-1xx245*K)=# #1.32xx10^-2*mol.# Why did I change the temperature to #"degrees Kelvin"#?

Avery Adams · Answer

undefined To find the molar quantity of the gas, you can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L atm / K mol)
T = temperature (in Kelvin)

First, convert the pressure from mmHg to atm and the temperature from Celsius to Kelvin:

732.6 mmHg * (1 atm / 760 mmHg) = 0.96447 atm
-28.2°C + 273.15 = 244.95 K

Now, plug the values into the ideal gas law equation:

0.96447 atm * 0.275 L = n * 0.0821 L atm / K mol * 244.95 K

Solve for n:

n = (0.96447 atm * 0.275 L) / (0.0821 L atm / K mol * 244.95 K)
n ≈ 0.01129 mol

So, the molar quantity of the gas is approximately 0.01129 mol.