At `10^(@)C`, the value of the density of a fixed mass of an ideal gas divided by its pressure is x. at `110^(@)C`, this ratio is

Options

A. x
B. `(383)/(283) x`
C. `(10)/(110) x`
D. `(283)/(383) x`

Answer

To solve the problem, we need to find the ratio of the density of a fixed mass of an ideal gas divided by its pressure at two different temperatures. We are given that at \(10^\circ C\), this ratio is \(x\), and we need to find this ratio at \(110^\circ C\).

### Step-by-step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the 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 in Kelvin.

2. **Express Density in Terms of Mass and Volume**:
The density \(\rho\) of the gas can be expressed as:
\[ \rho = \frac{m}{V} \]
where \(m\) is the mass of the gas.

3. **Substitute Density into the Ideal Gas Law**:
Rearranging the ideal gas law gives us:
\[ V = \frac{nRT}{P} \]
Substituting this into the density equation:
\[ \rho = \frac{m}{V} = \frac{mP}{nRT} \]

4. **Find the Ratio \(\frac{\rho}{P}\)**:
Dividing both sides by \(P\):
\[ \frac{\rho}{P} = \frac{m}{nRT} \]
This shows that \(\frac{\rho}{P}\) is directly proportional to \(\frac{1}{T}\):
\[ \frac{\rho}{P} \propto \frac{1}{T} \]

5. **Calculate the Temperatures in Kelvin**:
Convert the temperatures from Celsius to Kelvin:
- For \(10^\circ C\):
\[ T_1 = 10 + 273 = 283 \, K \]
- For \(110^\circ C\):
\[ T_2 = 110 + 273 = 383 \, K \]

6. **Set Up the Proportionality**:
Using the proportionality established:
\[ \frac{\rho_1}{P_1} = x \quad \text{at } T_1 = 283 \, K \]
\[ \frac{\rho_2}{P_2} = x' \quad \text{at } T_2 = 383 \, K \]
From the proportionality, we have:
\[ \frac{x'}{x} = \frac{T_1}{T_2} = \frac{283}{383} \]

7. **Solve for \(x'\)**:
Rearranging gives:
\[ x' = x \cdot \frac{283}{383} \]

8. **Final Result**:
Therefore, the ratio of density to pressure at \(110^\circ C\) is:
\[ x' = \frac{283}{383} x \]