A balloon containing an ideal gas has a volume of `10 litre` and temperature of `17^(@)`. If it is heated slowly to `75^(@)C`, the work done by the gas inside the balloon is (neglect elasticity of the balloon and take atmospheric pressure as `10^(5)` Pa)

Explanation

To solve the problem of calculating the work done by the gas inside the balloon when it is heated from 17°C to 75°C, we can follow these steps:

### Step 1: Identify the given values
- Volume of the gas, \( V_i = 10 \, \text{litres} \)
- Initial temperature, \( T_i = 17^\circ C = 17 + 273 = 290 \, K \)
- Final temperature, \( T_f = 75^\circ C = 75 + 273 = 348 \, K \)
- Atmospheric pressure, \( P = 10^5 \, \text{Pa} \)

### Step 2: Convert the volume from litres to cubic meters
Since 1 litre = \( 10^{-3} \, m^3 \), we convert:
\[ V_i = 10 \, \text{litres} = 10 \times 10^{-3} \, m^3 = 0.01 \, m^3 \]

### Step 3: Use the ideal gas law to relate the initial and final states
From the ideal gas law, we know that:
\[ \frac{V_i}{V_f} = \frac{T_i}{T_f} \]
Thus, we can express \( V_f \) in terms of \( V_i \):
\[ V_f = V_i \times \frac{T_f}{T_i} \]

### Step 4: Calculate the final volume \( V_f \)
Substituting the values:
\[ V_f = 0.01 \, m^3 \times \frac{348 \, K}{290 \, K} \]
Calculating this gives:
\[ V_f \approx 0.01 \times 1.2 = 0.012 \, m^3 \]

### Step 5: Calculate the change in volume \( \Delta V \)
\[ \Delta V = V_f - V_i = 0.012 \, m^3 - 0.01 \, m^3 = 0.002 \, m^3 \]

### Step 6: Calculate the work done \( W \)
The work done by the gas at constant pressure is given by:
\[ W = P \Delta V \]
Substituting the values:
\[ W = 10^5 \, \text{Pa} \times 0.002 \, m^3 = 200 \, J \]

### Final Answer
The work done by the gas inside the balloon is:
\[ \boxed{200 \, J} \]

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