At constant pressure what would be the percentage decrease in the density of an ideal gas for a 10% increase in the temperature.

Options

A. 0.1
B. 0.091
C. 0.11
D. 0.1209

Answer

To solve the problem of finding the percentage decrease in the density of an ideal gas for a 10% increase in temperature at constant pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: We know from the ideal gas law that \( PV = nRT \). Rearranging this, we can express density (\( D \)) in terms of temperature (\( T \)):
\[ D = \frac{PM}{RT} \]
where \( P \) is the pressure, \( M \) is the molar mass, and \( R \) is the universal gas constant.

2. **Establish Initial Conditions**: Let the initial temperature be \( T_1 \) and the initial density be \( D_1 \). Therefore, we can write:
\[ D_1 = \frac{PM}{RT_1} \]

3. **Increase in Temperature**: If the temperature is increased by 10%, the new temperature \( T_2 \) becomes:
\[ T_2 = T_1 + 0.1T_1 = 1.1T_1 \]

4. **Calculate New Density**: The new density \( D_2 \) at the new temperature \( T_2 \) can be expressed as:
\[ D_2 = \frac{PM}{RT_2} = \frac{PM}{R(1.1T_1)} = \frac{D_1}{1.1} \]

5. **Finding the Percentage Decrease in Density**: The percentage decrease in density can be calculated using the formula:
\[ \text{Percentage Decrease} = \frac{D_1 - D_2}{D_1} \times 100 \]
Substituting \( D_2 \):
\[ \text{Percentage Decrease} = \frac{D_1 - \frac{D_1}{1.1}}{D_1} \times 100 = \left(1 - \frac{1}{1.1}\right) \times 100 \]

6. **Simplifying the Expression**: Calculating the fraction:
\[ 1 - \frac{1}{1.1} = 1 - 0.9090 = 0.0909 \]
Thus,
\[ \text{Percentage Decrease} = 0.0909 \times 100 = 9.09\% \]

### Final Answer:
The percentage decrease in the density of the ideal gas for a 10% increase in temperature at constant pressure is approximately **9.09%**.