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 (ρ) and pressure (P) of an ideal gas at two different temperatures. We are given that the ratio at 10°C (283 K) is 5, and we need to find the ratio at 110°C (383 K).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The ratio of density to pressure at 10°C (T1 = 283 K) is given as:
\[ \frac{\rho}{P} = R_1 = 5 \]
- The temperature at 10°C in Kelvin:
\[ T_1 = 10 + 273 = 283 \text{ K} \]
- The temperature at 110°C in Kelvin:
\[ T_2 = 110 + 273 = 383 \text{ K} \]

2. **Use the Ideal Gas Law:**
The ideal gas law states:
\[ P = \frac{\rho RT}{M} \]
Rearranging gives:
\[ \frac{\rho}{P} = \frac{RT}{M} \]
This shows that the ratio of density to pressure is directly proportional to the temperature (T).

3. **Set Up the Proportionality:**
Since we know that:
\[ \frac{\rho}{P} \propto T \]
We can write the ratios at the two temperatures:
\[ \frac{R_1}{R_2} = \frac{T_1}{T_2} \]

4. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ \frac{5}{R_2} = \frac{283}{383} \]

5. **Cross-Multiply to Solve for R2:**
Cross-multiplying gives:
\[ 5 \cdot 383 = R_2 \cdot 283 \]
\[ 1915 = R_2 \cdot 283 \]
Now, solving for \( R_2 \):
\[ R_2 = \frac{1915}{283} \approx 6.76 \]

6. **Final Result:**
Thus, the ratio of density to pressure at 110°C is approximately:
\[ R_2 \approx 6.76 \]

### Conclusion:
The ratio of density to pressure of the ideal gas at 110°C is approximately **6.76**.