A container of gas with a volume of 952 mL at a temperature of 321 K exerted a pressure of 108 atm. Calculate the pressure of this same amount of gas in a 1.236 L container at a temperature of 59 oF.
To solve this problem, we will use the combined gas law, which relates the pressure, volume, and temperature of a gas when its amount (number of moles) stays constant. The formula is:
P1V1/T1 = P2V2/T2
where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature. We can use any units for pressure, volume, and temperature as long as we use the same units consistently for each measurement.
Let's start by converting the initial temperature to Kelvin, since the combined gas law requires absolute temperature. We add 273.15 to the temperature in Celsius:
T1 = 321 + 273.15 = 594.15 K
Next, we can apply the combined gas law to find the initial number of moles of gas:
P1V1/T1 = P2V2/T2
n1 = (P1V1/T1) × (T2/V2) / P2
n1 = (108 atm × 0.952 L / 594.15 K) × (59 oF + 459.67) / 1.236 L / (nearly the same as 59 oF + 459.67 = 518.67 R) / atm
n1 = 0.0476 mol
Now we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of an ideal gas. The formula is:
PV = nRT
where R is the gas constant, which has the value of 0.08206 L·atm/mol·K. We need to convert the final temperature to Kelvin as well:
T2 = (59 oF + 459.67) × 5/9 = 288.15 K
Now we can solve for the final pressure:
P2 = n1RT2/V2
P2 = (0.0476 mol × 0.08206 L·atm/mol·K × 288.15 K) / 1.236 L
P2 = 1.11 atm
Therefore, the pressure of the gas in a 1.236 L container at a temperature of 59 oF is about 1.11 atm, which is significantly lower than the initial pressure of 108 atm due to the larger volume and lower temperature.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's calculate the number of moles of gas in the initial container:
V1 = 952 mL = 0.952 L (since 1 L = 1000 mL)
P1 = 108 atm
T1 = 321 K
Now, we can use the ideal gas law equation to find n:
n = (P1 * V1) / (R * T1)
n = (108 atm * 0.952 L) / (0.0821 L.atm/mol.K * 321 K)
n ≈ 0.368 mol
Next, we need to find the pressure of the gas in the new container. We can rearrange the ideal gas law equation to solve for P2:
P2 = (n * R * T2) / V2
V2 = 1.236 L
T2 = 59 oF = (59 - 32) oC = (59 - 32) K ≈ 26.85 K
P2 = (0.368 mol * 0.0821 L.atm/mol.K * 26.85 K) / 1.236 L
P2 ≈ 1.604 atm
Therefore, the pressure of the same amount of gas in a 1.236 L container at a temperature of 59 oF is approximately 1.604 atm.