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

### Step 1: Understand the relationship between pressure and volume
The problem states that pressure (P) is inversely proportional to the square of the volume (V). This can be expressed mathematically as:
\[ P \propto \frac{1}{V^2} \]
This implies:
\[ P = \frac{k}{V^2} \]
where \( k \) is a constant.

### Step 2: Relate pressure and volume to temperature using the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
Substituting the expression for pressure from Step 1 into the ideal gas law:
\[ \frac{k}{V^2} \cdot V = nRT \]
This simplifies to:
\[ \frac{k}{V} = nRT \]

### Step 3: Rearranging the equation
From the equation \( \frac{k}{V} = nRT \), we can rearrange it to find a relationship between temperature (T) and volume (V):
\[ TV = \frac{k}{nR} \]
This shows that \( TV \) is a constant.

### Step 4: Analyze the effect of increasing temperature
Since \( TV = \text{constant} \), if the temperature (T) increases, the volume (V) must decrease to keep the product \( TV \) constant. This means:
- If \( T \) increases, then \( V \) decreases.

### Step 5: Determine the change in volume
When the volume decreases, the change in volume (\( dV \)) will be negative:
\[ dV = V_f - V_i \]
where \( V_f < V_i \), hence \( dV < 0 \).

### Step 6: Calculate the work done by the gas
The work done by the gas is given by:
\[ W = P \, dV \]
Since pressure (P) is always positive, and \( dV \) is negative, the work done will be:
\[ W < 0 \]
This indicates that the work done by the gas is negative.

### Conclusion
Therefore, when the temperature of the gas increases, the work done by the gas is negative.

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

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