Answer:
volume = 0.11 m³
Explanation:
The question tells us that 4.30 mol of oxygen gas is at a pressure of 99.7 kPa and at a temperature of 35.0 °C. The question then asks us to calculate the volume of the gas.
To do this, we have to use the ideal gas equation:
[tex]\boxed{pV = nRT}[/tex],
where:
• p = pressure in Pa
• V = volume in m³
• n = number of moles
• R = molar gas constant (8.31 JK⁻¹mol⁻¹)
• T = temperature in K
Since the pressure and temperature are given in kPa and °C respectively, we have to convert them to the units required by the formula:
Now we simply have to substitute the values into the equation and solve for V :
pV = nRT
⇒ 99.7 × 10³ × V = 4.30 × 8.31 × 308
⇒ V = [tex]\frac{4.30 \times 8.31 \times 308}{99.7 \times 10^3}[/tex]
⇒ V = 0.11 m³
Therefore, the volume of the gas is 0.11 m³.
According to the ideal gas law, 4.30 molecule of o2 occupies 0.11 m³ at 99.7 kPa with 35.0 °C.
The relationship between a gas's pressure P, volumes V, and t in the range between low pressures and extremely high temperatures, when the gas's water moves virtually independently of one another, is known as the ideal gas, also known as the perfect gas law.
PV = nRT
p = pressure in Pa
V = volume in m³
n = number of moles
R = molar gas constant (8.31 JK⁻¹mol⁻¹)
T = temperature in K
pressure ⇒ 99.7 × 10³ Pa
temperature ⇒ 35.0 + 273 = 308 K
pV=nRT
= 99.7×10³×V
= 4.30×8.31×308
V=(4.30×8.31×308)/(99.7×10³3 )
V= 0.11 m³
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