If a gas at 1.2 atm of pressure and 22.0 degrees of Celsius occupies 0.350L, what pressure will hold the same sample of gas in a volume of 0.050L if the temperature of the gas increases 25 degrees of Celsius?
To find the new pressure, we can use the combined gas law, which states that the ratio of the initial pressure to the final pressure is equal to the ratio of the initial volume to the final volume, multiplied by the ratio of the final temperature to the initial temperature.
Mathematically, the combined gas law can be expressed as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (1.2 atm)
V1 = initial volume (0.350 L)
T1 = initial temperature (22.0 degrees Celsius + 273.15) -> converted to Kelvin
P2 = final pressure (unknown)
V2 = final volume (0.050 L)
T2 = final temperature (22.0 degrees Celsius + 25 degrees Celsius + 273.15) -> converted to Kelvin
Let's plug in the values and solve for P2:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
(1.2 atm * 0.350 L) / (22.0 + 273.15 K) = (P2 * 0.050 L) / (22.0 + 25 + 273.15 K)
Simplifying the equation:
0.42 / 295.15 = (P2 * 0.050) / 320.15
Cross-multiplying:
0.42 * 320.15 = P2 * 0.050
Solving for P2:
P2 = (0.42 * 320.15) / 0.050
P2 ≈ 2.691 atm
Therefore, the new pressure that will hold the same sample of gas in a volume of 0.050 L when the temperature increases by 25 degrees Celsius is approximately 2.691 atm.
To solve this problem, we first need to apply the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
Step 1: Convert temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature.
Given initial temperature = 22.0°C
Add 273.15 to 22.0 to get the temperature in Kelvin:
22.0 + 273.15 = 295.15 K
Given increased temperature = 25.0°C
Add 273.15 to 25.0 to get the increased temperature in Kelvin:
25.0 + 273.15 = 298.15 K
Step 2: Solve for the initial number of moles of gas
Given the initial pressure = 1.2 atm
Given the initial volume = 0.350 L
Given the initial temperature in Kelvin = 295.15 K
Rearranging the ideal gas law equation, we can solve for the number of moles (n):
n = PV / RT
n = (1.2 atm * 0.350 L) / (0.0821 L*atm/(mol*K) * 295.15 K)
Calculating:
n = (0.42 atm*L) / (24.29 L*mol^-1*K^-1)
n ≈ 0.0173 mol
Step 3: Solve for the final pressure
Given the final volume = 0.050 L
Given the increased temperature in Kelvin = 298.15 K
The number of moles (n) remains the same for the gas sample.
We can use the rearranged ideal gas law equation to solve for the final pressure:
P2 = (nRT2) / V2
P2 = (0.0173 mol * 0.0821 L*atm/(mol*K) * 298.15 K) / 0.050 L
Calculating:
P2 ≈ 0.173 atm
Therefore, the final pressure required to hold the same sample of gas in a volume of 0.050 L when the temperature increased by 25 degrees Celsius is approximately 0.173 atm.