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`

Explanation

To solve the problem, we need to find the ratio of the density of a fixed mass of an ideal gas to its pressure at two different temperatures. Let's break it down step by step.

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin

### Step 2: Relate Density to Pressure and Temperature
Density (\( \rho \)) can be expressed as:
\[ \rho = \frac{m}{V} \]
Where \( m \) is the mass of the gas. From the ideal gas law, we can express density in terms of pressure and temperature:
\[ \rho = \frac{P \cdot M}{R \cdot T} \]
Where \( M \) is the molar mass of the gas.

### Step 3: Find the Ratio of Density to Pressure
From the above equation, we can derive the ratio of density to pressure:
\[ \frac{\rho}{P} = \frac{M}{R \cdot T} \]
This means that the ratio of density to pressure is directly proportional to the molar mass and inversely proportional to the temperature.

### Step 4: Calculate the Ratio at Two Different Temperatures
Let \( x \) be the ratio at \( 10^\circ C \):
\[ x = \frac{M}{R \cdot T_1} \]
Where \( T_1 = 10 + 273 = 283 \, K \).

Now, at \( 110^\circ C \), let the new ratio be \( x' \):
\[ x' = \frac{M}{R \cdot T_2} \]
Where \( T_2 = 110 + 273 = 383 \, K \).

### Step 5: Express the New Ratio in Terms of \( x \)
We can relate \( x' \) to \( x \):
\[ \frac{x'}{x} = \frac{T_1}{T_2} \]
Substituting the values of \( T_1 \) and \( T_2 \):
\[ \frac{x'}{x} = \frac{283}{383} \]
Thus, we can express \( x' \) as:
\[ x' = x \cdot \frac{283}{383} \]

### Conclusion
The ratio of the density of the gas to its pressure at \( 110^\circ C \) is:
\[ x' = \frac{283}{383} \cdot x \]