Question

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

°C?

Answer 1

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
`(O_2)` if it occupies 31.6 mL at
`91^(o)C`.

Step-by-step explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

`1 mL = 0.001 L\\31.6 mL = 31.6 mL * (0.001 L)/(1 mL)\\= 0.0316 L`

Temperature =
`91^(o)C = (91 + 273) K = 364 K`

As molar mass of
`O_2` is 32 g/mol. Hence, the number of moles of
`O_2` are calculated as follows.

`No. of moles = (mass)/(molar mass)\\= (0.023 g)/(32 g/mol)\\= 0.00072 mol`

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

`PV = nRT\\P * 0.0316 L = 0.00072 mol * 0.0821 L atm/mol K * 364 K\\P = (0.00072 mol * 0.0821 L atm/mol K * 364 K)/(0.0316 L)\\= 0.681 atm`

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
`(O_2)` if it occupies 31.6 mL at
`91^(o)C`.