Question

Suppose that the atmospheric temperature and pressure at the top of a ski run are −5°C and 713 mmHg, respectively. At the bottom of the run, the temperature and pressure are 0°C and 734 mmHg, respectively. How many more moles of oxygen does a skier take in with a lungful of air at the bottom of the run than at the top? Express your answer as a percentage.

Answer 1

At the top of the run, there are approximately 0.0448 moles of oxygen, while at the bottom, there are approximately 0.0462 moles. The difference is approximately 0.0014 moles or 3.12%.

Step-by-step explanation:

To calculate the number of moles of oxygen at the top and bottom of the ski run, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, let's calculate the number of moles of oxygen at the top of the ski run:

• Temperature (T1) = -5°C = 268 K
• Pressure (P1) = 713 mmHg = 95.0 kPa

Using the equation PV = nRT, we can rearrange it to n = PV / RT. Plugging in the values, we get:

n1 = (95.0 kPa) / (8.314 J/(mol·K) * 268 K) ≈ 0.0448 mol

Next, let's calculate the number of moles of oxygen at the bottom of the ski run:

• Temperature (T2) = 0°C = 273 K
• Pressure (P2) = 734 mmHg = 97.9 kPa

Again, using the equation PV = nRT and plugging in the values, we get:

n2 = (97.9 kPa) / (8.314 J/(mol·K) * 273 K) ≈ 0.0462 mol

Finally, to calculate the difference in moles, we subtract n1 from n2:

Difference = n2 - n1 ≈ 0.0462 mol - 0.0448 mol ≈ 0.0014 mol

To express the difference as a percentage, we divide the difference by n1 and multiply by 100:

Percentage Difference = (0.0014 mol / 0.0448 mol) * 100 ≈ 3.12%