find the dry volume at STP of 500 mL of nitrogen gas measured over water at 15 degree C and 680 mmHg. the water vapor exerts a partial pressure of 12.8 torr at this temperature ( vapor pressure).

(P1V1/T1) = (P2V2/T2)

P1 = 680-12.8 = ?
T1 = 15 + 273
V1 = 500 mL
P2, and T2 are STP. Solve for V2

To find the dry volume of nitrogen gas at STP (Standard Temperature and Pressure), we need to correct for the presence of water vapor.

Step 1: Convert the temperature from Celsius to Kelvin
To do this, we add 273 to the Celsius temperature.
T(K) = T(°C) + 273
T(K) = 15 + 273
T(K) = 288 K

Step 2: Convert the pressure from mmHg to atm
To convert from mmHg to atm, we divide by 760 (since 760 mmHg is equal to 1 atm).
P(atm) = P(mmHg) / 760
P(atm) = 680 mmHg / 760
P(atm) ≈ 0.895 atm

Step 3: Calculate the partial pressure of water vapor in atm
To convert the vapor pressure from torr to atm, we divide by 760.
Pvapor(atm) = Pvapor(torr) / 760
Pvapor(atm) = 12.8 torr / 760
Pvapor(atm) ≈ 0.0168 atm

Step 4: Correct the total pressure for the presence of water vapor
To find the pressure of the dry gas, we subtract the partial pressure of water vapor from the total pressure.
Pdry(atm) = P(atm) - Pvapor(atm)
Pdry(atm) = 0.895 atm - 0.0168 atm
Pdry(atm) ≈ 0.8782 atm

Step 5: Apply the ideal gas law to find the volume
The ideal gas law equation is:
PV = nRT

We have:
Pdry(atm) = 0.8782 atm
V = 500 mL (since this is the volume of nitrogen gas measured over water)
n = ? (unknown)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = 288 K (temperature in Kelvin)

Rearranging the ideal gas law equation to solve for n:
n = Pdry(atm) * V / (R * T)

Substituting the given values:
n = 0.8782 atm * (500 mL / 1000) / (0.0821 L·atm/(mol·K) * 288 K)
n ≈ 0.0206 moles

Step 6: Calculate the dry volume
Since the volume of a mole of gas at STP is 22.4 L, we can calculate the dry volume.
Vdry = n * 22.4 L/mol
Vdry = 0.0206 moles * 22.4 L/mol
Vdry ≈ 0.4614 L

Therefore, the dry volume of the nitrogen gas at STP is approximately 0.4614 liters.

To find the dry volume of nitrogen gas at STP (standard temperature and pressure), we need to subtract the volume occupied by water vapor from the total volume of the gas over water. Here's how you can calculate it:

1. Convert the given atmospheric pressure from mmHg to torr.
680 mmHg = 680 torr

2. Convert the given temperature from Celsius to Kelvin.
15 degrees C = 15 + 273.15 K = 288.15 K

3. Calculate the partial pressure of water vapor.
Since the vapor pressure of water at 15 degrees C is given as 12.8 torr, the partial pressure of water vapor is also 12.8 torr.

4. Subtract the partial pressure of water vapor from the atmospheric pressure to get the pressure of the dry nitrogen gas.
Dry nitrogen gas pressure = Atmospheric pressure - Partial pressure of water vapor
Dry nitrogen gas pressure = 680 torr - 12.8 torr = 667.2 torr

5. Apply the ideal gas law to calculate the dry volume of nitrogen gas.
V₁ / P₁ = V₂ / P₂

V₁ = Initial volume of nitrogen gas over water = 500 mL = 500 cm³
P₁ = Initial pressure of nitrogen gas over water = Atmospheric pressure = 680 torr
V₂ = Final volume of dry nitrogen gas at STP (unknown)
P₂ = Final pressure of dry nitrogen gas at STP = 1 atm = 760 torr

V₂ = (V₁ * P₂) / P₁
V₂ = (500 cm³ * 760 torr) / 680 torr

6. Convert the volume from cm³ to mL.
V₂ = 500 * 760 / 680 mL
V₂ = 558.82 mL

Therefore, the dry volume of nitrogen gas at STP, measured over water at 15 degrees C and 680 mmHg, is approximately 558.82 mL.