Question

Suppose you have a 0.500-L cylinder that contains 0.150 mol of oxygen gas, O2, 25 degree celsius. C. how would the pressure inside the cylinder change if you added 0.150 mol of sulfur dioxide gas, SO2, to the oxygen already in the cylinder

Answer 1

The pressure inside the cylinder will change to 14.686 atm

Step-by-step explanation

Given:

Volume, V = 0.500 L

Moles of O2 = 0.150 mol

Temperature, T = 25 °C = (25 + 273.15 K) = 298.15 K

Step-by-step solution:

The first step is to calculate the pressure exerted by O2 using ideal gas equation:

`\begin{gathered} PV=nRT \\ \end{gathered}`

Plugging the values of the parameters into the formula:

`\begin{gathered} P*0.500\text{ }L=0.150\text{ }mol*0.0821\text{ }atm•L/mol•K*298.15K \\ \\ P=\frac{3.67171725\text{ }atm.L}{0.500L}=7.343\text{ }atm \end{gathered}`

The pressure exerted by O2 = 7.343 atm

If 0.150 mol of sulfur dioxide, SO2 was added to the oxygen gas already in the cylinder, then the pressure inside the cylinder will change by a factor of 2. That is:

Total pressure will be = 2 x 7.343 atm = 14.686 atm

Or the total pressure can be calculated using the ideal equation above, and putting n = 0.150 mol + 0.150 mol = 0.3 mol

`\begin{gathered} P*0.500\text{ }L=0.30\text{ }mol*0.0821\text{ }atm•L/mol•K*298.15K \\ \\ P=\frac{7.3434345\text{ }atm.L}{0.500L}=14.686\text{ }atm \end{gathered}`