#0.250*mol# quantities of dinitrogen and sulfur dioxide are enclosed in a #2.5*L# vessel at a temperature of #300*K#. What are the partial pressures of each gas, and what is the total pressure?

Question

#0.250*mol# quantities of dinitrogen and sulfur dioxide are enclosed in a #2.5*L# vessel at a temperature of #300*K#.  What are the partial pressures of each gas, and what is the total pressure?

Jeremiah Adams · Accepted Answer

#"We use Dalton's Law of Partial Pressures..........."# to get #P_"Total"=5*atm#

In a gaseous mixture, the partial pressure exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures.

And thus we simply use the Ideal Gas Law:

#P=(nRT)/V#

#P_(SO_2)=(0.25*molxx0.0821*(L*atm)/(K*mol)xx300*K)/(2.5*L)#

#=2.5*atm#

And likewise, #P_(N_2)=P_(SO_2)=2.5*atm#

#P_"Total"=P_(SO_2)+P_(N_2)=??*atm. #

Natalie Anton · Answer

undefined To find the partial pressures of each gas and the total pressure, you can use the ideal gas law, $PV = nRT$, where:

- $P$ is the pressure,
- $V$ is the volume,
- $n$ is the number of moles,
- $R$ is the ideal gas constant, and
- $T$ is the temperature in Kelvin.

Given:
- $n_{\text{N}_2} = 0.250$ mol
- $n_{\text{SO}_2} = 0.250$ mol
- $V = 2.5$ L
- $T = 300$ K

First, find the total number of moles ($n_{\text{total}}$):
$$ n_{\text{total}} = n_{\text{N}_2} + n_{\text{SO}_2} $$

Then, substitute into the ideal gas law to find the total pressure ($P_{\text{total}}$):
$$ P_{\text{total}} = \frac{n_{\text{total}}RT}{V} $$

Next, use the mole fractions to find the partial pressures of each gas:
$$ P_{\text{N}_2} = \frac{n_{\text{N}_2}}{n_{\text{total}}} \times P_{\text{total}} $$
$$ P_{\text{SO}_2} = \frac{n_{\text{SO}_2}}{n_{\text{total}}} \times P_{\text{total}} $$

Substitute the given values into these equations to find the partial pressures and total pressure.