At the top of a mountain a thermometer reads `7^(0)C` and barometer reads `70 cm` of `Hg`. At the bottom of the mountain the barometer reads `76 cm` of `Hg` and thermometer reads `27^(0)C`. The density of air at the top of mountain is "_" times the density at the bottom.

Options

A. `0.80`
B. `0.98`
C. `0.99`
D. `0.97`

Explanation

To find the ratio of the density of air at the top of the mountain to the density of air at the bottom, we can use the ideal gas law and the relationship between pressure, temperature, and density.

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

2. **Express Density in Terms of Pressure and Temperature**:
We can express the number of moles \( n \) as:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is the molar mass. Thus, we can rewrite the ideal gas law as:
\[ PV = \frac{m}{M}RT \]
Dividing both sides by \( V \):
\[ P = \frac{m}{V} \cdot \frac{RT}{M} \]
Here, \( \frac{m}{V} \) is the density \( \rho \). Therefore, we can express density as:
\[ P = \rho \cdot \frac{RT}{M} \]
Rearranging gives us:
\[ \rho = \frac{PM}{RT} \]

3. **Set Up the Ratio of Densities**:
We want to find the ratio of the density at the top of the mountain (\( \rho_t \)) to the density at the bottom of the mountain (\( \rho_b \)):
\[ \frac{\rho_t}{\rho_b} = \frac{P_t}{P_b} \cdot \frac{T_b}{T_t} \]
where \( P_t \) and \( P_b \) are the pressures at the top and bottom, and \( T_t \) and \( T_b \) are the temperatures at the top and bottom respectively.

4. **Convert Temperatures to Kelvin**:
- At the top: \( T_t = 7^\circ C = 7 + 273 = 280 \, K \)
- At the bottom: \( T_b = 27^\circ C = 27 + 273 = 300 \, K \)

5. **Substitute the Given Values**:
- Pressure at the top: \( P_t = 70 \, cm \, Hg \)
- Pressure at the bottom: \( P_b = 76 \, cm \, Hg \)

Now substituting these values into the density ratio:
\[ \frac{\rho_t}{\rho_b} = \frac{70}{76} \cdot \frac{300}{280} \]

6. **Calculate the Ratio**:
First, calculate \( \frac{70}{76} \):
\[ \frac{70}{76} \approx 0.9211 \]
Next, calculate \( \frac{300}{280} \):
\[ \frac{300}{280} \approx 1.0714 \]
Now multiply these two results:
\[ \frac{\rho_t}{\rho_b} \approx 0.9211 \cdot 1.0714 \approx 0.987 \]

7. **Final Result**:
The density of air at the top of the mountain is approximately \( 0.98 \) times the density at the bottom.