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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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Most popular questions from this chapter

Show that, for an ideal gas, $$ P=\frac{1}{3} \rho v_{\mathrm{rms}}^{2} $$ where \(P\) is the pressure, \(\rho\) is the mass density, and \(v_{\mathrm{rms}}\) is the rms speed of the gas molecules.
(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?
Find the molar mass of ammonia (NH \(_{3}\) ).
For divers going to great depths, the composition of the air in the tank must be modified. The ideal composition is to have approximately the same number of \(\mathrm{O}_{2}\) molecules per unit volume as in surface air (to avoid oxygen poisoning), and to use helium instead of nitrogen for the remainder of the gas (to avoid nitrogen narcosis, which results from nitrogen dissolving in the bloodstream). Of the molecules in dry surface air, \(78 \%\) are \(\mathrm{N}_{2}, 21 \%\) are \(\mathrm{O}_{2},\) and \(1 \%\) are \(\mathrm{Ar}\). (a) How many \(\mathrm{O}_{2}\) molecules per \(\mathrm{m}^{3}\) are there in surface air at \(20.0^{\circ} \mathrm{C}\) and 1.00 atm? (b) For a diver going to a depth of \(100.0 \mathrm{m}\) what percentage of the gas molecules in the tank should be \(\mathrm{O}_{2} ?\) (Assume that the density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\) and the temperature is $20.0^{\circ} \mathrm{C} .$ )
If 2.0 mol of nitrogen gas \(\left(\mathrm{N}_{2}\right)\) are placed in a cubic box, \(25 \mathrm{cm}\) on each side, at 1.6 atm of pressure, what is the rms speed of the nitrogen molecules?
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