Question

ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. how many ozone molecules are present in 2.00 l of air if the temperature is 269 k temperature and pressure is 1.01×10⁻³ atm?
r = 0.081 (\frac{lcdot atm}{molcdot k}) or 8.314 (\frac{kpacdot l}{molcdot k})
options:

• 5.51×10¹⁹ molecules
• 9.15×10⁻⁵ molecules
• 1.82×10²⁰ molecules
• 1.09×10⁴ molecules

Explanation:

Step1: Apply Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant, and \( T \) is temperature. We need to solve for \( n \): \( n=\frac{PV}{RT} \). Given \( P = 1.01\times10^{-3}\ atm \), \( V = 2.00\ L \), \( R = 0.0821\ \frac{L\cdot atm}{mol\cdot K} \), \( T = 269\ K \).

Substitute values: \( n=\frac{(1.01\times10^{-3}\ atm)(2.00\ L)}{(0.0821\ \frac{L\cdot atm}{mol\cdot K})(269\ K)} \)

Calculate denominator: \( 0.0821\times269 \approx 22.0849 \)

Calculate numerator: \( 1.01\times10^{-3}\times2.00 = 2.02\times10^{-3} \)

Then \( n=\frac{2.02\times10^{-3}}{22.0849} \approx 9.15\times10^{-5}\ mol \)

Step2: Convert Moles to Molecules

Use Avogadro's number (\( N_A = 6.022\times10^{23}\ molecules/mol \)): \( N = n\times N_A \)

Substitute \( n = 9.15\times10^{-5}\ mol \): \( N = 9.15\times10^{-5}\ mol\times6.022\times10^{23}\ molecules/mol \approx 5.51\times10^{19}\ molecules \)

Answer:

5.51×10¹⁹ molecules (Corresponding option: 5.51x10¹⁹ molecules)