Questions: On the summit of Mt. Everest, the atmospheric pressure is 210 mmHg and the air density is 0.429 g / L. Assume that the molar mass of air is 29.0 g / mol. Calculate the temperature (in °C) at the summit.

On the summit of Mt. Everest, the atmospheric pressure is 210 mmHg and the air density is 0.429 g / L. Assume that the molar mass of air is 29.0 g / mol. Calculate the temperature (in °C) at the summit.
Transcript text: On the summit of Mt. Everest, the atmospheric pressure is 210. mmHg and the air density is $0.429 \mathrm{~g} / \mathrm{L}$. Assume that the molar mass of air is $29.0 \mathrm{~g} / \mathrm{mol}$. Calculate the temperature (in ${ }^{\circ} \mathrm{C}$ ) at the summit. $\qquad$ ${ }^{\circ} \mathrm{C}$
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Solution

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Solution Steps

Step 1: Identify the Given Values

We are given the following values:

  • Atmospheric pressure, \( P = 210 \, \text{mmHg} \)
  • Air density, \( \rho = 0.429 \, \text{g/L} \)
  • Molar mass of air, \( M = 29.0 \, \text{g/mol} \)
Step 2: Convert Pressure to Standard Units

Convert the pressure from mmHg to atmospheres (atm) using the conversion factor \( 1 \, \text{atm} = 760 \, \text{mmHg} \):

\[ P = \frac{210 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.2763 \, \text{atm} \]

Step 3: Use the Ideal Gas Law

The ideal gas law is given by:

\[ PV = nRT \]

We can rearrange this to solve for temperature \( T \):

\[ T = \frac{PV}{nR} \]

Step 4: Relate Density to Moles

The number of moles \( n \) can be expressed in terms of density \( \rho \), volume \( V \), and molar mass \( M \):

\[ n = \frac{\rho V}{M} \]

Substitute \( n \) into the ideal gas law:

\[ T = \frac{PV}{\left(\frac{\rho V}{M}\right) R} = \frac{PM}{\rho R} \]

Step 5: Substitute Known Values

Substitute the known values into the equation:

\[ T = \frac{(0.2763 \, \text{atm})(29.0 \, \text{g/mol})}{(0.429 \, \text{g/L})(0.0821 \, \text{L·atm·K}^{-1}\text{·mol}^{-1})} \]

Step 6: Calculate the Temperature in Kelvin

Perform the calculation:

\[ T = \frac{(0.2763)(29.0)}{(0.429)(0.0821)} = \frac{8.0127}{0.0352} = 227.6 \, \text{K} \]

Step 7: Convert Temperature to Celsius

Convert the temperature from Kelvin to Celsius:

\[ T_{\text{C}} = T_{\text{K}} - 273.15 = 227.6 - 273.15 = -45.55 \, ^\circ \text{C} \]

Final Answer

\(\boxed{-45.55 \, ^\circ \text{C}}\)

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