Question

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

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 $91^{o}C$.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

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.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

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.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

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 $91^{o}C$.