Respuesta :
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen [tex](O_2)[/tex] if it occupies 31.6 mL at [tex]91^{o}C[/tex].
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
[tex]1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L[/tex]
Temperature = [tex]91^{o}C = (91 + 273) K = 364 K[/tex]
As molar mass of [tex]O_2[/tex] is 32 g/mol. Hence, the number of moles of [tex]O_2[/tex] are calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol[/tex]
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.
[tex]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[/tex]
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen [tex](O_2)[/tex] if it occupies 31.6 mL at [tex]91^{o}C[/tex].
