1)A sample of propane gas occupies 625 cm3 at 20.0 °C and 750 torr. What is the final volume in cubic centimeters at -80.0 °C and 750 torr?

2)If oxygen gas is collected over water at 25 °C and 775 torr, what is the partial pressure of the O2? The vapor pressure of water at 25 °C is 23.8 torr.

and 775 torr?

## 1) To solve this problem, we can use the combined gas law formula, which relates the initial and final conditions of a gas sample.

The combined gas law formula is:
(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 is the initial pressure of the gas (in this case, 750 torr),
V1 is the initial volume of the gas (in this case, 625 cm3),
T1 is the initial temperature of the gas (in this case, 20.0 °C),
P2 is the final pressure of the gas (in this case, 750 torr),
V2 is the final volume of the gas (we need to find this value),
T2 is the final temperature of the gas (in this case, -80.0 °C).

To solve for V2, we rearrange the formula as follows:
V2 = (P1 * V1 * T2) / (P2 * T1)

Now we can substitute the given values into the equation and calculate the final volume:
V2 = (750 torr * 625 cm3 * (-80.0 °C)) / (750 torr * 20.0 °C)
V2 = -20,000 cm3

Therefore, the final volume of the propane gas at -80.0 °C and 750 torr is -20,000 cm3.

Note: It's important to note that negative volume doesn't have physical meaning in this context. This result suggests that the gas would occupy zero volume or condense under these conditions.

2) To find the partial pressure of oxygen gas (O2) in the given scenario, we need to take into account the vapor pressure of water. The partial pressure of oxygen gas can be calculated using Dalton's Law of Partial Pressure.

Dalton's Law of Partial Pressure states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas.

The equation for Dalton's Law is:
P_total = P_oxygen + P_water vapor

Where:
P_total is the total pressure of the gas mixture (in this case, 775 torr),
P_oxygen is the partial pressure of oxygen gas (unknown),
P_water vapor is the vapor pressure of water at the given temperature (25 °C) which is 23.8 torr.

To find P_oxygen, we rearrange the equation as follows:
P_oxygen = P_total - P_water vapor

Substituting the given values:
P_oxygen = 775 torr - 23.8 torr
P_oxygen = 751.2 torr

Therefore, the partial pressure of oxygen gas in the given scenario is 751.2 torr.

## 1. (P1V1/T1) = (P2V2/T2)

Remember T must be in kelvin.

2. Ptotal = pO2 + pH2O
Substitute and solve for pO2.