Question

A sample of 0.400 L Ne initially at 1.181 atm and 240.ºC and a 0.150 L sample of O₂ initially at 0.794 atm and 130.ºC are added to 500. mL flask held at 25.0ºC. Which choice is closest to the total pressure in the flask?

Enter your answer to three significant figures in units of atm.

Answer 1

Total pressure = 0.725 atm

Explanation

According to ideal gas law,

moles of gas = [(pressure) * (volume)] / [(R) * (temperature)]

For Ne,

pressure = 1.181 atm

volume = 0.400 L

temperature = 240 ^oC = 513 K

moles Ne = [(pressure Ne) * (volume Ne)] / [(R) * (temperature Ne)]

moles Ne = [(1.181 atm) * (0.400 L)] / [(0.0821 L-atm/mol-K) * (513 K)]

moles Ne = 0.01121889 mol

Similarly, moles O₂ = 0.0036 mol

Total moles = moles Ne + moles O₂

Total moles = (0.01121889 mol) + (0.0036 mol)

Total moles = 0.0148 mol

Total pressure = [(Total moles) * (R) * (temperature)] / (volume)

Total pressure = [(0.0148 mol) * (0.0821 L-atm/mol-K) * (298 K)] / (0.500 L)

Total pressure = 0.725 atm