Liquid water can be separated into hydrogen gas and oxygen gas through electrolysis. 1 mole of hydrogen gas and 0.5 moles of oxygen gas is produced 1 mole of liquid water collected in separate 10L containers at 1 atm. Will the temperatures of the gases be equal? Use the ideal gas law.
To determine whether the temperatures of the gases will be equal, we can use the ideal gas law equation: PV = nRT.
Given that 1 mole of hydrogen gas is produced in a 10L container at 1 atm, we can calculate the hydrogen gas's temperature:
PV = nRT
(1 atm)(10L) = (1 mole)(R)(T)
Rearranging the equation to solve for T, we get:
T = (PV) / (nR)
Plugging in the values:
T = (1 atm)(10L) / (1 mole)(R)
Now, given that 0.5 moles of oxygen gas are produced in another separate 10L container at 1 atm, we can calculate the oxygen gas's temperature using the same equation:
T = (PV) / (nR)
T = (1 atm)(10L) / (0.5 moles)(R)
Since R is a constant, we can see that both the hydrogen gas and oxygen gas will have the same temperature, as T is directly proportional to PV and inversely proportional to n.
Therefore, the temperatures of the gases will be equal.
To determine whether the temperatures of the gases will be equal, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (1 atm),
V is the volume (10 L),
n is the number of moles of gas, and
R is the ideal gas constant (0.0821 L·atm/(mol·K)).
T represents the temperature in Kelvin.
We know that 1 mole of hydrogen gas and 0.5 moles of oxygen gas is produced from 1 mole of liquid water and collected in separate 10L containers.
Let's calculate the temperature of the hydrogen gas first:
n_H2 = 1 mole
P = 1 atm
V = 10 L
R = 0.0821 L·atm/(mol·K)
Rearranging the equation, we get:
T_H2 = PV / (n_H2 * R)
T_H2 = (1 atm * 10L) / (1 mol * 0.0821 L·atm/(mol·K))
T_H2 = 10 / 0.0821 K
T_H2 ≈ 121.7 K
Now, let's calculate the temperature of the oxygen gas:
n_O2 = 0.5 moles
P = 1 atm
V = 10 L
R = 0.0821 L·atm/(mol·K)
Rearranging the equation, we get:
T_O2 = PV / (n_O2 * R)
T_O2 = (1 atm * 10L) / (0.5 mol * 0.0821 L·atm/(mol·K))
T_O2 = 20 / 0.0821 K
T_O2 ≈ 243.1 K
Based on the calculations, the temperatures of the gases will not be equal. The temperature of the hydrogen gas (T_H2) is approximately 121.7 K, while the temperature of the oxygen gas (T_O2) is approximately 243.1 K.
To determine if the temperatures of the gases will be equal, we can use the ideal gas law, which states:
PV = nRT
where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- R is the ideal gas constant
- T is the temperature of the gas
Since the gases are collected in separate 10L containers, the volume of both gases is 10L. We are given that 1 mole of hydrogen gas and 0.5 moles of oxygen gas are produced, so n for hydrogen gas is 1 mol and n for oxygen gas is 0.5 mol.
Assuming the pressure and volume are constant for both gases, we can rearrange the ideal gas law to solve for temperature:
T = PV / (nR)
Let's calculate the temperatures of hydrogen and oxygen gases separately:
For hydrogen gas:
PH2 × VH2 = nH2 × RH2 × TH2
For oxygen gas:
PO2 × VO2 = nO2 × RO2 × TO2
Since the pressure and volume are the same for both gases, we can simplify the equations:
PH2 × 10L = 1 mol × RH2 × TH2
PO2 × 10L = 0.5 mol × RO2 × TO2
Divide both equations by their respective number of moles:
PH2 × 10L / 1 mol = RH2 × TH2
PO2 × 10L / 0.5 mol = RO2 × TO2
Simplify further:
PH2 × 10L = RH2 × TH2
PO2 × 20L = RO2 × TO2
Since the pressure and volume for both gases are the same, we can equate their temperatures:
RH2 × TH2 = RO2 × TO2
Therefore, the temperatures of the hydrogen and oxygen gases will be equal.