a sample of a gas occupies a volume of 1025 mL at 75 degrees Celsius and 0.75 atm. What will be the new volume if temperature decreases to 35 degrees Celsius and pressure increases to 1.25 atm
To solve this problem, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (0.75 atm)
V1 = initial volume (1025 mL)
T1 = initial temperature in Kelvin (75 degrees Celsius + 273.15 = 348.15 K)
P2 = final pressure (1.25 atm)
V2 = final volume (unknown)
T2 = final temperature in Kelvin (35 degrees Celsius + 273.15 = 308.15 K)
Let's substitute the values into the equation:
(0.75 * 1025) / (348.15) = (1.25 * V2) / (308.15)
Now, we can solve for V2:
(0.75 * 1025 * 308.15) / (348.15) = 1.25 * V2
V2 = (0.75 * 1025 * 308.15) / (1.25 * 348.15)
V2 ≈ 666.43 mL
Therefore, the new volume of the gas would be approximately 666.43 mL when the temperature decreases to 35 degrees Celsius and pressure increases to 1.25 atm.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of gas,
R is the ideal gas constant, and
T is the temperature of the gas in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T₁ = 75°C + 273.15 = 348.15 K
T₂ = 35°C + 273.15 = 308.15 K
Next, we can rearrange the ideal gas law equation to solve for the new volume (V₂):
V₂ = (nR * T₂) / P₂
To find the number of moles (n), we can use the equation:
n = PV / RT
Now, let's calculate n using the initial conditions (P₁, V₁, and T₁):
n = (P₁ * V₁) / (R * T₁)
Since the conditions (P₂, V₂, and T₂) are changing, we will use the same value for n but plug in the new values of P₂ and T₂:
V₂ = (nR * T₂) / P₂
Let's plug in the values and calculate:
Given:
V₁ = 1025 mL = 1025 cm³ = 1025 * 10^-3 dm³ = 1.025 dm³
P₁ = 0.75 atm
T₁ = 348.15 K
P₂ = 1.25 atm
T₂ = 308.15 K
R is the ideal gas constant, which is approximately 0.0821 L·atm/(K·mol).
n = (P₁ * V₁) / (R * T₁)
n = (0.75 atm * 1.025 dm³) / (0.0821 L·atm/(K·mol) * 348.15 K)
n ≈ 0.0286 mol
V₂ = (nR * T₂) / P₂
V₂ = ((0.0286 mol) * (0.0821 L·atm/(K·mol)) * 308.15 K) / 1.25 atm
V₂ ≈ 0.605 L
Therefore, the new volume (V₂) will be approximately 0.605 liters.