At standard temperature and pressure (0∘C and 1.00atm), 1.00mol of an ideal gas occupies a volume of 22.4L. What volume would the same amount of gas occupy at the same pressure and 30∘C?
To solve this problem, we can use the combined gas law equation:
(P₁V₁) / (T₁) = (P₂V₂) / (T₂)
Where:
P₁ = initial pressure (1.00 atm)
V₁ = initial volume (22.4 L)
T₁ = initial temperature (0°C + 273.15 = 273.15 K)
P₂ = final pressure (1.00 atm)
V₂ = final volume (unknown)
T₂ = final temperature (30°C + 273.15 = 303.15 K)
Substituting the values into the equation, we have:
(1.00 atm * 22.4 L) / (273.15 K) = (1.00 atm * V₂) / (303.15 K)
Simplifying the equation, we get:
22.4 L / 273.15 K = V₂ / 303.15 K
Cross-multiplying and solving for V₂, we have:
V₂ = (22.4 L * 303.15 K) / 273.15 K
V₂ = 24.79 L
Therefore, the volume that the same amount of gas would occupy at 30°C and 1.00 atm pressure is approximately 24.79 L.
To solve this problem, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure
V = volume
n = moles
R = ideal gas constant
T = temperature
The ideal gas constant (R) is a constant value that depends on the units of the other variables used in the equation. For this problem, we can use the value of R as 0.0821 L·atm/(mol·K).
We are given that the initial volume (V₁) is 22.4 L, the initial temperature (T₁) is 0 °C (which we need to convert to Kelvin), the final temperature (T₂) is 30 °C (which we also need to convert to Kelvin), and the final pressure (P₂) is the same as the initial pressure (P₁) at 1.00 atm.
Now, let's calculate the final volume (V₂):
First, convert the initial and final temperatures to Kelvin:
T₁ = 0 °C + 273.15 = 273.15 K (initial temperature)
T₂ = 30 °C + 273.15 = 303.15 K (final temperature)
Now, we can rearrange the ideal gas law equation to solve for V₂:
V₂ = (nRT₂) / P₂
We know that n = 1.00 mol, R = 0.0821 L·atm/(mol·K), T₂ = 303.15 K, and P₂ = 1.00 atm.
Plugging in these values:
V₂ = (1.00 mol * 0.0821 L·atm/(mol·K) * 303.15 K) / 1.00 atm
By calculating this expression, we find:
V₂ ≈ 24.8 L
Therefore, the volume of the gas at the same pressure and 30 °C would be approximately 24.8 L.
(V1/T1) = (V2/T2)
Remember T must be in kelvin.
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