To solve this problem, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas.
The combined gas law equation is:
P₁V₁/T₁ = P₂V₂/T₂
Where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
We are given the initial volume and pressure, as well as the standard pressure. Since the temperature is not given and remains constant, we can simplify the combined gas law equation as:
P₁V₁ = P₂V₂
Substituting the given values:
745.0 mmHg × 500.0 mL = 760 mmHg × V₂
Solving for V₂:
V₂ = (745.0 mmHg × 500.0 mL) / 760 mmHg
V₂ = 389.8 mL (rounded to the nearest tenth)
Therefore, the volume of the gas at standard pressure would be 389.8 mL.
Answers & Comments
To solve this problem, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas.
The combined gas law equation is:
P₁V₁/T₁ = P₂V₂/T₂
Where:
P₁ = initial pressure
V₁ = initial volume
T₁ = initial temperature
P₂ = final pressure
V₂ = final volume
T₂ = final temperature
We are given the initial volume and pressure, as well as the standard pressure. Since the temperature is not given and remains constant, we can simplify the combined gas law equation as:
P₁V₁ = P₂V₂
Substituting the given values:
745.0 mmHg × 500.0 mL = 760 mmHg × V₂
Solving for V₂:
V₂ = (745.0 mmHg × 500.0 mL) / 760 mmHg
V₂ = 389.8 mL (rounded to the nearest tenth)
Therefore, the volume of the gas at standard pressure would be 389.8 mL.