A flashbulb of volume 2.40 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 18.0 °C. How many grams of O2(g) does the flashbulb contain? g

Question

A flashbulb of volume 2.40 mL contains O₂(g) at a pressure of 2.30 atm and a temperature of 18.0 °C. How many grams of O₂(g) does the flashbulb contain?

     g   

Answer 1

Hello, Destiny,

Santon D. has the correct solution in the comments. Use the Ideal Gas Law: PV = nRT, where n is the number of moles of the gas. The temperature must be in degrees Kelvin and the gas constant, R, needs to be in units compatible with the input data. I picked R to be 0.082958 L*atm/(K*mole). That way we only need to convert the 2.4 ml into liters by dividing by 1000 (0.0024 liters). Convert 18.0 C to the temperature in Kelvin by adding 273.15.

Rearrange the ideal gas law to isolate n, the number of moles:

n = PV/RT

Have your data in a table, clearly marked, and then enter the values.

n = 2.30atm)(0.0024L)/(0.0821L*atm/(K*mole))*(291.15K)

Check to see if the units all cancel to leave only moles and then solve.

I found 0.000231 moles in a quick calculation. To convert that into grams, multiply the moles by oxygen's molar mass of 32 g/mole, to yield 0.00739 grams of O₂

Seems too low to me, so please check the calculation,

Bob