A sample of an ideal gas has a volume of 3.20 L at 14.80 °C and 1.50 atm. What is the volume of the gas at 21.80 °C and 0.989 atm?
the volume is directly proportional to the (absolute) temperature (ºK)
and inversely proportional to the pressure
v = 3.20 L * (1.50 / .989) * [(21.80 + 273) / (14.80 + 273)]
To solve this problem, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law equation is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (1.50 atm)
V1 = initial volume (3.20 L)
T1 = initial temperature (14.80 °C + 273.15) in Kelvin
P2 = final pressure (0.989 atm)
V2 = final volume (unknown)
T2 = final temperature (21.80 °C + 273.15) in Kelvin
Now, let's plug in the given values into the equation and solve for V2.
(1.50 atm * 3.20 L) / (14.80 °C + 273.15 K) = (0.989 atm * V2) / (21.80 °C + 273.15 K)
(4.80 atm * L) / (288.95 K) = (0.989 atm * V2) / (295.95 K)
Cross multiplying, we get:
(4.80 atm * L) * (295.95 K) = (0.989 atm * V2) * (288.95 K)
1419.576 atm * L * K = 285.765 atm * V2
Dividing both sides by 285.765 atm:
1419.576 atm * L * K / 285.765 atm = V2
Simplifying the expression:
V2 = 4.96 L
Therefore, the volume of the gas at 21.80 °C and 0.989 atm is approximately 4.96 L.
To solve this problem, we can use the combined gas law formula, which is derived from Boyle's law, Charles's law, and Avogadro's law.
The combined gas law formula is:
(P₁ * V₁) / (T₁) = (P₂ * V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures,
V₁ and V₂ are the initial and final volumes,
T₁ and T₂ are the initial and final temperatures.
Let's assign the given values to the variables:
P₁ = 1.50 atm
V₁ = 3.20 L
T₁ = 14.80 °C = 273.95 K (convert to Kelvin using the formula: K = °C + 273.15)
P₂ = 0.989 atm
V₂ = ?
T₂ = 21.80 °C = 295.95 K
Now we can plug the values into the combined gas law formula and solve for V₂:
(1.50 atm * 3.20 L) / (273.95 K) = (0.989 atm * V₂) / (295.95 K)
To find V₂, we can cross multiply and then divide:
(1.50 atm * 3.20 L) * (295.95 K) = (0.989 atm) * V₂ * (273.95 K)
Simplifying the equation:
1440.3 atm*L*K = 270.97 atm * V₂
Dividing by 270.97 atm:
V₂ = (1440.3 atm*L*K) / (270.97 atm)
Calculating V₂:
V₂ ≈ 5.32 L
Therefore, the volume of the gas at 21.80 °C and 0.989 atm is approximately 5.32 L.