A sample of 0.010 mol of oxygen gas is confined at 37 °C and 0.216 atmosphere. What would be the pressure of this sample at 15 °C with the same volume?
the pressure is proportional to the absolute (Kelvin) temperature
0.216 / (37 + 273) = p / (15 + 273)
.201
To find the pressure of the sample at 15°C, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature (in Kelvin)
First, we need to convert temperatures from Celsius to Kelvin:
37 °C = 273 + 37 = 310 K (original temperature)
15 °C = 273 + 15 = 288 K (final temperature)
Now, we can rearrange the ideal gas law equation to solve for the pressure:
P1V1/T1 = P2V2/T2
Substituting the given values:
P1 = 0.216 atm (initial pressure)
V1 = ? (volume, unchanged)
T1 = 310 K (initial temperature)
P2 = ? (final pressure)
V2 = ? (volume, unchanged)
T2 = 288 K (final temperature)
Simplifying the equation:
P1/T1 = P2/T2
Substituting the values:
(0.216 atm) / (310 K) = P2 / (288 K)
Now, solve for P2:
P2 = (0.216 atm) * (288 K) / (310 K)
P2 ≈ 0.2 atm
Therefore, the pressure of the sample at 15 °C with the same volume would be approximately 0.2 atm.
To determine the pressure of the sample at 15 °C, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (0.216 atm)
V1 = initial volume (unknown)
T1 = initial temperature (37 °C + 273.15 K)
P2 = final pressure (unknown)
V2 = final volume (unknown, assumed to be the same as the initial volume)
T2 = final temperature (15 °C + 273.15 K)
Rearranging the equation to solve for P2, we have:
P2 = (P1 * V1 * T2) / (V2 * T1)
Now we need to determine the initial volume V1. Since the sample contains 0.010 mol of oxygen gas, we can use the ideal gas law equation to calculate the initial volume. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (0.216 atm)
V = volume (unknown, assumed to be the same as the final volume)
n = number of moles of gas (0.010 mol)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (37 °C + 273.15 K)
Rearranging the equation to solve for V, we have:
V = (n * R * T) / P
Now we can substitute the values into the equations. First, calculate the initial volume V1:
V1 = (0.010 mol * 0.0821 L·atm/mol·K * (37 °C + 273.15 K)) / 0.216 atm
Next, calculate the final pressure P2:
P2 = (0.216 atm * V1 * (15 °C + 273.15 K)) / (V1 * (37 °C + 273.15 K))
Simplifying the equation:
P2 = (0.216 atm * (15 °C + 273.15 K)) / (37 °C + 273.15 K)
Calculating the values:
P2 = (0.216 atm * 288.15 K) / 310.15 K
P2 ≈ 0.201 atm
Therefore, the pressure of the sample at 15 °C with the same volume would be approximately 0.201 atm.