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Answer :
First, we determine the number of moles of each gas. For hydrogen gas, the number of moles is calculated using the formula
[tex]$$
\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2}.
$$[/tex]
Given that the mass of [tex]$H_2$[/tex] is [tex]$2.00$[/tex] g and its molar mass is [tex]$2.00$[/tex] g/mol, we have
[tex]$$
\text{moles of } H_2 = \frac{2.00}{2.00} = 1.0 \text{ mol}.
$$[/tex]
Next, for nitrogen gas,
[tex]$$
\text{moles of } N_2 = \frac{\text{mass of } N_2}{\text{molar mass of } N_2}.
$$[/tex]
With a mass of [tex]$8.00$[/tex] g and a molar mass of [tex]$28.00$[/tex] g/mol for nitrogen,
[tex]$$
\text{moles of } N_2 = \frac{8.00}{28.00} \approx 0.2857 \text{ mol}.
$$[/tex]
The total number of moles in the vessel is the sum of the moles of each gas:
[tex]$$
n_{\text{total}} = 1.0 + 0.2857 \approx 1.2857 \text{ mol}.
$$[/tex]
Now, we use the ideal gas law to calculate the pressure. The ideal gas law is given by
[tex]$$
PV = nRT,
$$[/tex]
where:
- [tex]$P$[/tex] is the pressure,
- [tex]$V$[/tex] is the volume,
- [tex]$n$[/tex] is the number of moles,
- [tex]$R$[/tex] is the gas constant, and
- [tex]$T$[/tex] is the temperature in Kelvin.
Rearranging for [tex]$P$[/tex], we get
[tex]$$
P = \frac{nRT}{V}.
$$[/tex]
Substitute the known values: [tex]$n = 1.2857 \text{ mol}$[/tex], [tex]$R = 0.08206 \, \text{L atm/(mol K)}$[/tex], [tex]$T = 273 \text{ K}$[/tex], and [tex]$V = 10 \text{ L}$[/tex]:
[tex]$$
P = \frac{1.2857 \times 0.08206 \times 273}{10}.
$$[/tex]
Carrying out the multiplication and division gives
[tex]$$
P \approx 2.88 \text{ atm}.
$$[/tex]
Thus, the pressure exerted by the gas mixture in the vessel is approximately
[tex]$$
\boxed{2.88 \text{ atm}}.
$$[/tex]
[tex]$$
\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2}.
$$[/tex]
Given that the mass of [tex]$H_2$[/tex] is [tex]$2.00$[/tex] g and its molar mass is [tex]$2.00$[/tex] g/mol, we have
[tex]$$
\text{moles of } H_2 = \frac{2.00}{2.00} = 1.0 \text{ mol}.
$$[/tex]
Next, for nitrogen gas,
[tex]$$
\text{moles of } N_2 = \frac{\text{mass of } N_2}{\text{molar mass of } N_2}.
$$[/tex]
With a mass of [tex]$8.00$[/tex] g and a molar mass of [tex]$28.00$[/tex] g/mol for nitrogen,
[tex]$$
\text{moles of } N_2 = \frac{8.00}{28.00} \approx 0.2857 \text{ mol}.
$$[/tex]
The total number of moles in the vessel is the sum of the moles of each gas:
[tex]$$
n_{\text{total}} = 1.0 + 0.2857 \approx 1.2857 \text{ mol}.
$$[/tex]
Now, we use the ideal gas law to calculate the pressure. The ideal gas law is given by
[tex]$$
PV = nRT,
$$[/tex]
where:
- [tex]$P$[/tex] is the pressure,
- [tex]$V$[/tex] is the volume,
- [tex]$n$[/tex] is the number of moles,
- [tex]$R$[/tex] is the gas constant, and
- [tex]$T$[/tex] is the temperature in Kelvin.
Rearranging for [tex]$P$[/tex], we get
[tex]$$
P = \frac{nRT}{V}.
$$[/tex]
Substitute the known values: [tex]$n = 1.2857 \text{ mol}$[/tex], [tex]$R = 0.08206 \, \text{L atm/(mol K)}$[/tex], [tex]$T = 273 \text{ K}$[/tex], and [tex]$V = 10 \text{ L}$[/tex]:
[tex]$$
P = \frac{1.2857 \times 0.08206 \times 273}{10}.
$$[/tex]
Carrying out the multiplication and division gives
[tex]$$
P \approx 2.88 \text{ atm}.
$$[/tex]
Thus, the pressure exerted by the gas mixture in the vessel is approximately
[tex]$$
\boxed{2.88 \text{ atm}}.
$$[/tex]
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