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Chapter 13: Problem 53

A tank of compressed air of volume \(1.0 \mathrm{m}^{3}\) is pressurized to 20.0 atm at \(T=273 \mathrm{K}\). A valve is opened and air is released until the pressure in the tank is 15.0 atm. How many air molecules were released?

Short Answer

Expert verified
Answer: Approximately 1.35 x 10^26 air molecules are released.

Step by step solution

01

Write down the Ideal Gas Law equation

The Ideal Gas Law equation is: PV = nRT where: P is the pressure (in atm) V is the volume (in m^3) n is the number of moles R is the ideal gas constant (0.0821 L·atm/mol·K) T is the temperature (in K)
02

Calculate initial number of moles

Using the initial pressure, volume, and temperature, calculate the initial number of moles: P1 = 20.0 atm V1 = 1.0 m^3 = 1000 L (1 m^3 = 1000 L) T1 = 273 K n1 = (P1 * V1) / (R * T1) Insert the given values and then solve for n1: n1 = (20.0 atm * 1000 L) / (0.0821 L·atm/mol·K * 273 K) n1 ≈ 895.64 mol
03

Calculate final number of moles

Using the final pressure, initial volume, and initial temperature, calculate the final number of moles: P2 = 15.0 atm n2 = (P2 * V1) / (R * T1) Insert the given values and then solve for n2: n2 = (15.0 atm * 1000 L) / (0.0821 L·atm/mol·K * 273 K) n2 ≈ 671.73 mol
04

Find the difference in moles

To find the number of moles released, subtract the final number of moles from the initial number of moles: Δn = n1 - n2 Δn ≈ 223.91 mol
05

Calculate the number of air molecules released

Using Avogadro's constant (6.022 x 10^23 molecules/mol), we can calculate the number of air molecules released: Number of air molecules released = Δn * Avogadro's constant Number of air molecules released ≈ 223.91 mol * 6.022 x 10^23 molecules/mol Number of air molecules released ≈ 1.35 x 10^26 molecules The number of air molecules released is approximately 1.35 x 10^26 molecules.

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