a gas occupies a volume of 50.0mL at degrees Celsius and 630mmHg.at what temperature in degrees Celsius would the pressure be 101.3kPa if the volume remains constant?
(p1/t1) = (p2/t1).
Substitute and solve. You don't have t1 listed.
1865.74 = t2
To find the temperature in degrees Celsius at which the pressure would be 101.3 kPa, we can use the combined gas law equation, which states that:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
where P₁ and T₁ are the initial pressure and temperature, P₂ is the final pressure (101.3 kPa in this case), V₁ is the initial volume, V₂ is the final volume (which remains constant), and T₂ is the final temperature that we want to find.
Given:
P₁ = 630 mmHg
V₁ = 50.0 mL
P₂ = 101.3 kPa
Let's first convert the initial pressure and volume to the same unit as the final pressure:
1 kPa = 7.50062 mmHg (approximately)
P₁ = 630 mmHg
P₁ = 630 / 7.50062 ≈ 84.0 kPa (approximately)
Now we can substitute the values into the combined gas law equation and solve for T₂:
(84.0 kPa × 50.0 mL) / T₁ = (101.3 kPa × 50.0 mL) / T₂
Since the volume remains constant, we can cancel out the V₁ and V₂ terms:
84.0 kPa / T₁ = 101.3 kPa / T₂
Cross-multiplying:
(84.0 kPa) × (T₂) = (101.3 kPa) × (T₁)
Now we can rearrange the equation to solve for T₂:
T₂ = (101.3 kPa × T₁) / 84.0 kPa
Substituting the known values, with T₁ being the initial temperature in degrees Celsius, we can find the final temperature in degrees Celsius.