Answer:
He have higher pressure at room temperature.
Explanation:
It is given that both the gases are kept in 5 L chambers.
Therefore, volume is constant.
Also, they both are at room temperature, so temperature is also constant.
Now, number of moles of [tex]O_2[/tex] = [tex]\dfrac{Given\ mass}{Molecular \ mass}=\dfrac{5}{32}=0.16\ mol.[/tex]
Also, number of moles of He =[tex]\dfrac{Given\ mass}{Molecular \ mass}=\dfrac{5}{4}=1.25\ mol.[/tex]
Now, according to GAS LAW,
[tex]PV=nRT[/tex] ( all terms have their usual meaning).
In this case, V, R and T are constant.
So, pressure is directly proportional to n i.e number of moles.
So, moles of He is more than moles of [tex]O_2[/tex].
Therefore, He have higher pressure at room temperature.
Hence , this is the required solution.
The gas that has a higher pressure is He.
Number of moles of oxygen gas = 5.00 g /32 g/mol = 0.156 moles
From PV = nRT
P = ?
V = 5 L
n = 0.156 moles
T = 25 + 273 = 298 K
R = 0.082 atmLK-1mol-1
P = nRT/V
P = 0.156 moles × 0.082 atmLK-1mol-1 × 298 K/5 L
P = 0.76 atm
Number of moles of He = 5/4 g/mol = 1.25 moles
P = ?
V = 5 L
n = 1.25 moles
T = 25 + 273 = 298 K
R = 0.082 atmLK-1mol-1
P = nRT/V
P =1.25 moles × 0.082 atmLK-1mol-1 × 298 K/5 L
P = 6.11 atm
The gas that has a higher pressure is He.
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