Question

1. today’s pressure is 29.71 inches of mercury. standard pressure is 29.92 inches of mercury. standard pressure is also equivalent to 1.00 atmospheres of pressure. by volume, dry air is 20.95% oxygen. the molecular weight of oxygen is 32.00 g/mol. using the ideal gas law below, determine how many grams of oxygen are in the room.

pv = grt/mw
p must be in atm
v must be in l
r= 0.08206 l-atm/mol-k
t must be in kelvins (°c + 273.15 = k)
handwritten: volume = 2327, kelvins = 293.15

Explanation:

Step1: Convert pressure to atm

First, we need to find the pressure \( P \) in atmospheres. We know that standard pressure (29.92 inches of mercury) is 1.00 atm. Today's pressure is 29.71 inches of mercury. So we can set up a proportion:
\( P = \frac{29.71}{29.92} \times 1.00 \, \text{atm} \)
Calculating that, \( P \approx 0.993 \, \text{atm} \)

Step2: Identify known values

We know:

• \( P \approx 0.993 \, \text{atm} \)
• \( V = 2327 \, \text{L} \) (from the handwritten note)
• \( R = 0.08206 \, \text{L-atm/mol-K} \)
• \( T = 293.15 \, \text{K} \) (from the handwritten note)
• \( MW = 32.00 \, \text{g/mol} \) (molecular weight of oxygen)
• Also, oxygen is 20.95% by volume, so we need to consider the volume of oxygen, \( V_{O_2} = 0.2095 \times V \)

First, calculate \( V_{O_2} \):
\( V_{O_2} = 0.2095 \times 2327 \, \text{L} \approx 487.5 \, \text{L} \)

Step3: Rearrange ideal gas law for \( g \)

The ideal gas law is \( PV = \frac{gRT}{MW} \). Rearranging for \( g \) (mass of oxygen), we get:
\( g = \frac{PV \times MW}{RT} \)

Now, substitute the values for oxygen (using \( V_{O_2} \) as \( V \) here, since we're calculating for oxygen):
\( g = \frac{0.993 \, \text{atm} \times 487.5 \, \text{L} \times 32.00 \, \text{g/mol}}{0.08206 \, \text{L-atm/mol-K} \times 293.15 \, \text{K}} \)

Step4: Calculate numerator and denominator

First, calculate the numerator:
\( 0.993 \times 487.5 \times 32.00 \approx 0.993 \times 15600 \approx 15500.8 \)

Then, calculate the denominator:
\( 0.08206 \times 293.15 \approx 24.05 \)

Step5: Divide numerator by denominator

\( g \approx \frac{15500.8}{24.05} \approx 644.5 \, \text{g} \) (approximate value, depending on rounding during steps)

Wait, let's do the calculation more precisely:

First, \( V_{O_2} = 0.2095 \times 2327 = 2327 \times 0.2095 \)
2327 0.2 = 465.4, 2327 0.0095 = 22.1065, so total \( V_{O_2} = 465.4 + 22.1065 = 487.5065 \, \text{L} \)

Now, numerator: \( 0.993 \times 487.5065 \times 32.00 \)
0.993 * 487.5065 ≈ 484.094
484.094 * 32.00 ≈ 15491.008

Denominator: \( 0.08206 \times 293.15 \)
0.08206 * 293.15 ≈ 24.053

Now, \( g = \frac{15491.008}{24.053} \approx 644.0 \, \text{g} \) (more precise)

Wait, maybe I made a mistake in the pressure conversion? Let's check the pressure conversion again. Standard pressure is 29.92 inHg = 1 atm. So today's pressure is 29.71 inHg. So \( P = \frac{29.71}{29.92} \times 1 \, \text{atm} \). Let's calculate that: 29.71 / 29.92 ≈ 0.993, that's correct.

Alternatively, maybe the volume of the room is 2327 L? The handwritten note says Volume = 2327, Kelvins = 293.15. So that's correct.

So putting it all together:

\( g = \frac{P \times V_{O_2} \times MW}{R \times T} \)

Substituting:

\( P = 0.993 \, \text{atm} \)

\( V_{O_2} = 0.2095 \times 2327 = 487.5065 \, \text{L} \)

\( MW = 32.00 \, \text{g/mol} \)

\( R = 0.08206 \, \text{L-atm/mol-K} \)

\( T = 293.15 \, \text{K} \)

So:

\( g = \frac{0.993 \times 487.5065 \times 32.00}{0.08206 \times 293.15} \)

Calculate step by step:

First, multiply 0.993 and 487.5065: 0.993 * 487.5065 ≈ 484.094

Then multiply by 32.00: 484.094 * 32 = 15491.008

Denominator: 0.08206 * 293.15 ≈ 24.053

Now divide: 15491.008 / 24.053 ≈ 644 grams (approximate)

Answer:

Approximately \(\boxed{644}\) grams (the exact value may vary slightly depending on rounding during calculations)