The ratio of the density and pressure of a fixed mass of an ideal gas is "5" at 10°C. Then this ratio at 110°C is

Options

A. 3.69
B. 5
C. 8.46
D. 6.76

Answer

To solve the problem, we need to find the ratio of density to pressure of a fixed mass of an ideal gas at two different temperatures. We know that the ratio of density (ρ) to pressure (P) at 10°C is 5, and we want to find this ratio at 110°C.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The temperature at 10°C in Kelvin is:
\[ T_1 = 10 + 273 = 283 \, \text{K} \]
- The temperature at 110°C in Kelvin is:
\[ T_2 = 110 + 273 = 383 \, \text{K} \]

2. **Use the Ideal Gas Law**:
- The ideal gas law states that:
\[ PV = nRT \]
- Rearranging this gives us:
\[ P = \frac{m}{V} \cdot \frac{R}{M} \cdot T \]
- Here, \(\frac{m}{V}\) is the density (ρ), so we can express pressure as:
\[ P = \rho \cdot \frac{R}{M} \cdot T \]

3. **Find the Ratio of Density to Pressure**:
- From the rearrangement, we can express the ratio of density to pressure as:
\[ \frac{\rho}{P} = \frac{M}{R \cdot T} \]
- This shows that the ratio \(\frac{\rho}{P}\) is inversely proportional to the temperature (T).

4. **Set Up the Ratios**:
- We can set up the ratio of the densities and pressures at the two temperatures:
\[ \frac{\rho_2}{P_2} = \frac{M}{R \cdot T_2} \]
- And we know:
\[ \frac{\rho_1}{P_1} = \frac{M}{R \cdot T_1} \]
- Therefore, we can write:
\[ \frac{\rho_2}{P_2} = \frac{T_1}{T_2} \cdot \frac{\rho_1}{P_1} \]

5. **Substituting Known Values**:
- We know that \(\frac{\rho_1}{P_1} = 5\) (given in the problem).
- Substituting the values:
\[ \frac{\rho_2}{P_2} = \frac{283}{383} \cdot 5 \]

6. **Calculate the Final Value**:
- Performing the calculation:
\[ \frac{\rho_2}{P_2} = \frac{283 \times 5}{383} = \frac{1415}{383} \approx 3.69 \]

### Conclusion:
The ratio of density to pressure at 110°C is approximately **3.69**.