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a gas mixture contains 0.150 mol of oxygen (o₂) gas, 0.419 mol of nitro…

Question

a gas mixture contains 0.150 mol of oxygen (o₂) gas, 0.419 mol of nitrogen (n₂) gas, and 0.211 mol of argon (ar) gas in a 0.250 l flask at 298 k. what is the total pressure of the mixture?
select the correct answer below:
20.6 atm
41.0 atm
76.3 atm
14.7 atm

Explanation:

Step1: Calculate total moles

Add moles of each gas.
$n_{total}=0.150 + 0.419+0.211$
$n_{total}=0.780$ mol

Step2: Use ideal gas law $PV = nRT$

Solve for $P$, where $R = 0.0821\ L\cdot atm/(mol\cdot K)$, $V=0.250\ L$, $T = 298\ K$, $n = 0.780$ mol.
$P=\frac{nRT}{V}$
$P=\frac{0.780\times0.0821\times298}{0.250}$
First calculate numerator: $0.780\times0.0821\times298 = 0.780\times24.4658=19.083324$
Then $P=\frac{19.083324}{0.250}=76.333296\approx76.3$ atm

Answer:

76.3 atm