In a process the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas increases, then work done by the gas:

Options

A. is positive
B. is negative
C. is zero
D. may be positive

Explanation

To solve the problem step by step, we will analyze the relationship between pressure, volume, and temperature of the gas, and then determine the work done by the gas when the temperature increases.

### Step 1: Understand the relationship between pressure and volume
Given that the pressure \( P \) of the gas is inversely proportional to the square of the volume \( V \), we can express this relationship mathematically as:
\[ P \propto \frac{1}{V^2} \]
This can be rewritten as:
\[ PV^2 = K \]
where \( K \) is a constant.

### Step 2: Relate pressure, volume, and temperature using the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
Substituting the expression for \( P \) from our earlier relationship into the ideal gas law gives:
\[ \left(\frac{K}{V^2}\right)V = nRT \]
This simplifies to:
\[ \frac{K}{V} = nRT \]

### Step 3: Rearranging the equation
Rearranging the equation, we find:
\[ TV = \frac{K}{nR} \]
This shows that the product of temperature \( T \) and volume \( V \) is constant.

### Step 4: Analyze the effect of increasing temperature
If the temperature \( T \) increases, to keep the product \( TV \) constant, the volume \( V \) must decrease. Therefore, we conclude:
\[ \text{If } T \text{ increases, then } V \text{ decreases.} \]

### Step 5: Determine the work done by the gas
The work done by the gas during a process is given by:
\[ W = P \Delta V \]
where \( \Delta V \) is the change in volume. Since we established that the volume decreases when the temperature increases, \( \Delta V \) will be negative.

### Step 6: Conclusion about the work done
Since pressure \( P \) is always positive and \( \Delta V \) is negative, the work done by the gas will be:
\[ W = P \Delta V < 0 \]
Thus, the work done by the gas is negative.

### Final Answer
The work done by the gas when the temperature increases is negative.

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