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Chapter 5: Problem 55

The steel reaction vessel of a bomb calorimeter, which has a volume of 75.0 \(\mathrm{mL}\) , is charged with oxygen gas to a pressure of 14.5 atm at \(22^{\circ} \mathrm{C}\) . Calculate the moles of oxygen in the reaction vessel.

Short Answer

Expert verified
The moles of oxygen gas in the reaction vessel can be calculated using the Ideal Gas Law Equation, PV = nRT. Given the volume V = 0.075 L, pressure P = 14.5 atm (converted to 1468712.5 Pa), and temperature T = 295 K, the moles of oxygen (n) can be calculated as: n = (1468712.5 Pa) × (0.075 L) / (8.314 J/molK × 295 K) n ≈ 0.0042 mol Therefore, there are approximately 0.0042 moles of oxygen gas in the reaction vessel.

Step by step solution

01

List the given values and convert all to SI units

Given: Volume, V = 75.0 mL = 0.075 L (1 L = 1000 mL) Pressure, P = 14.5 atm (1 atm = 101325 Pa, we'll convert later) Temperature, T = 22℃ = 295 K ( K = ℃ + 273)
02

Convert units

1. Pressure conversion 1 atm = 101325 Pa 14.5 atm = 14.5 × 101325 Pa = 1468712.5 Pa So now, P = 1468712.5 Pa
03

Rearrange the Ideal Gas Law Equation to solve for moles (n)

The Ideal Gas Law Equation is given by: PV = nRT We have to calculate the moles of oxygen gas, so we'll rearrange the equation to solve for n: n = PV / RT
04

Substitute the values and calculate n

Now, we can substitute the values of P, V, R, and T into the equation: n = (1468712.5 Pa) × (0.075 L) / (8.314 J/molK × 295 K) n ≈ 0.0042 mol So, there are approximately 0.0042 moles of oxygen gas in the reaction vessel.

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