If the pressure of an ideal gas at constant volume is decreased by 20% then the percentage change in temperature will be 

Options

A. 20% increase
B. 20% decrease
C. 80% increase
D. 80% decrease

Answer

To solve the problem, we will use the relationship between pressure and temperature for an ideal gas at constant volume. The equation governing the behavior of an ideal gas is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

Since the volume \( V \) and the number of moles \( n \) are constant in this scenario, we can simplify our analysis.

### Step 1: Understand the relationship between pressure and temperature
At constant volume, the pressure of an ideal gas is directly proportional to its temperature. This can be expressed mathematically as:

\[ P \propto T \]

This means that if the pressure changes, the temperature will change in the same proportion.

### Step 2: Calculate the new pressure
If the initial pressure is \( P_1 \), and it is decreased by 20%, the new pressure \( P_2 \) can be calculated as:

\[ P_2 = P_1 - 0.20 P_1 = 0.80 P_1 \]

### Step 3: Relate the change in pressure to the change in temperature
Since pressure is directly proportional to temperature, we can express the temperatures as:

- Initial temperature \( T_1 \) corresponding to pressure \( P_1 \)
- New temperature \( T_2 \) corresponding to pressure \( P_2 \)

Using the proportionality, we have:

\[ \frac{P_2}{P_1} = \frac{T_2}{T_1} \]

Substituting \( P_2 \):

\[ \frac{0.80 P_1}{P_1} = \frac{T_2}{T_1} \]

This simplifies to:

\[ 0.80 = \frac{T_2}{T_1} \]

### Step 4: Calculate the percentage change in temperature
From the above equation, we can express \( T_2 \) in terms of \( T_1 \):

\[ T_2 = 0.80 T_1 \]

Now, to find the percentage change in temperature, we calculate:

\[ \text{Percentage change in temperature} = \frac{T_1 - T_2}{T_1} \times 100 \]

Substituting \( T_2 \):

\[ \text{Percentage change in temperature} = \frac{T_1 - 0.80 T_1}{T_1} \times 100 \]

This simplifies to:

\[ = \frac{0.20 T_1}{T_1} \times 100 \]

\[ = 20\% \]

### Conclusion
Thus, if the pressure of an ideal gas at constant volume is decreased by 20%, the percentage change in temperature will also be a decrease of 20%.

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