Question

A 1.40 L sample of O2 at 645 Torr and 25 °C, and a 0.751 L sample of N2 at 1.13 atm and 25 °C, are both transferred to the same 2.00 L container at 25 °C. What are the partial pressures of the two gases and the total pressure inside the 2.00 L container? Show your work.

Answer 1

Answer:

• P(O₂) = 0.595 atm

• P(N₂) = 0.424 atm

• Total Pressure = 1.019 atm

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to calculate the moles of each one:

• O₂:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

• N₂:

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can calculate the partial pressure for each gas in the new container, because the number of moles did not change:

• O₂:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

• N₂:

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to calculate the total pressure:

• Pt = 0.595 atm+ 0.424 atm = 1.019 atm