A horizontal insulated cylinder is provided with frictionless non-conducting piston. On each side of the piston there is 50 litres of air at a pressure of 1 atmosphere and 273 K. Heat is slowly supplied to the air at the left hand side, until the piston has compressed the air on the right hand side to 2.5 atmosphere. Find Work done on the air on the right hand side

Answer

To solve the problem of finding the work done on the air on the right-hand side of the piston, we will follow these steps:

### Step 1: Understand the Initial Conditions
We start with 50 liters of air at a pressure of 1 atmosphere and a temperature of 273 K on both sides of the piston.

### Step 2: Convert Units
Convert the volume from liters to cubic meters for standard SI units:
\[ V = 50 \text{ liters} = 50 \times 10^{-3} \text{ m}^3 = 0.05 \text{ m}^3 \]

### Step 3: Determine the Final Conditions
The problem states that the pressure on the right side of the piston is compressed to 2.5 atmospheres. We need to convert this pressure to pascals:
\[ P_{\text{final}} = 2.5 \text{ atm} = 2.5 \times 101325 \text{ Pa} = 253312.5 \text{ Pa} \]

### Step 4: Calculate the Work Done
The work done on the gas can be calculated using the formula for work done on a gas during an isothermal process:
\[ W = P \Delta V \]
However, since the volume on the right side is compressed, we need to find the change in volume. Initially, the volume on the right side is 0.05 m³, and we need to find the final volume when the pressure is 2.5 atm.

Using the ideal gas law, we can find the final volume:
\[ PV = nRT \]
Assuming the number of moles (n) and the temperature (T) remain constant, we can set up the equation:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \(P_1 = 1 \text{ atm} = 101325 \text{ Pa}\)
- \(V_1 = 0.05 \text{ m}^3\)
- \(P_2 = 2.5 \text{ atm} = 253312.5 \text{ Pa}\)

Rearranging gives:
\[ V_2 = \frac{P_1 V_1}{P_2} = \frac{101325 \times 0.05}{253312.5} \approx 0.020 \text{ m}^3 \]

### Step 5: Calculate the Change in Volume
The change in volume (\(\Delta V\)) is:
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} = 0.020 - 0.05 = -0.030 \text{ m}^3 \]

### Step 6: Calculate Work Done
Now, substituting the values into the work done formula:
\[ W = P_{\text{final}} \Delta V = 253312.5 \times (-0.030) \approx -7599.375 \text{ J} \]
The work done on the gas is approximately 7599.375 J (the negative sign indicates work is done on the gas).

### Final Answer
The work done on the air on the right-hand side is approximately **7599.375 J**.

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