Question11) 0.7515 moles of nitrogen gas and 0.1135 moles of methane gas are placed in a 171.6 ml container at $20.8^{\circ} \mathrm{C}$ . What is the partial pressure (atm) of nitrogen gas? A) 1.14 B) 0.473 C) 16.0 D) 106 E) 226

Studdy Solution

STEP 1

1. We are given the moles of nitrogen gas and methane gas.
2. The volume of the container is 171.6 ml.
3. The temperature is given as $20.8^{\circ} \mathrm{C}$ .
4. We need to find the partial pressure of nitrogen gas in atm.
5. We will use the ideal gas law to find the partial pressure.

STEP 2

1. Convert the volume from ml to liters.
2. Convert the temperature from Celsius to Kelvin.
3. Use the ideal gas law to calculate the partial pressure of nitrogen gas.

STEP 3

Convert the volume from ml to liters.
$\text{Volume in liters} = \frac{171.6 \, \text{ml}}{1000} = 0.1716 \, \text{L}$

STEP 4

Convert the temperature from Celsius to Kelvin.
$\text{Temperature in Kelvin} = 20.8 + 273.15 = 293.95 \, \text{K}$

STEP 5

Use the ideal gas law to calculate the partial pressure of nitrogen gas. The ideal gas law is given by:
$PV = nRT$
Where: - $P$ is the pressure, - $V$ is the volume, - $n$ is the number of moles, - $R$ is the ideal gas constant ( $0.0821 \, \text{L atm/mol K}$ ), - $T$ is the temperature in Kelvin.
Rearrange the formula to solve for $P$ :
$P = \frac{nRT}{V}$
Substitute the values for nitrogen gas:
$P = \frac{0.7515 \, \text{moles} \times 0.0821 \, \text{L atm/mol K} \times 293.95 \, \text{K}}{0.1716 \, \text{L}}$
Calculate the partial pressure:
$P = \frac{0.7515 \times 0.0821 \times 293.95}{0.1716}$
$P \approx \frac{18.108}{0.1716}$
$P \approx 105.55 \, \text{atm}$
The closest answer choice is:
$\boxed{106}$

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