the pressure of a gas change linearly with volume from 10kPa, 200 cc to 50 kPa, 50 cc. (a) calculate the work done by the gas, (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas ?

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Calculate the Work Done by the Gas

1. **Identify the Points**: We have two points given:
- Point 1: Pressure \( P_1 = 10 \, \text{kPa} \) and Volume \( V_1 = 200 \, \text{cc} \) (which is \( 200 \times 10^{-6} \, \text{m}^3 \))
- Point 2: Pressure \( P_2 = 50 \, \text{kPa} \) and Volume \( V_2 = 50 \, \text{cc} \) (which is \( 50 \times 10^{-6} \, \text{m}^3 \))

2. **Convert Units**: Convert pressures from kPa to Pa:
- \( P_1 = 10 \times 10^3 \, \text{Pa} = 10,000 \, \text{Pa} \)
- \( P_2 = 50 \times 10^3 \, \text{Pa} = 50,000 \, \text{Pa} \)

3. **Draw the P-V Diagram**: Plot the points on a graph where the x-axis is Volume and the y-axis is Pressure. The points will form a trapezium.

4. **Calculate the Area Under the Curve**: The work done by the gas is equal to the area under the P-V curve. The area of a trapezium is given by:
\[ \text{Area} = \frac{1}{2} \times (P_1 + P_2) \times (V_2 - V_1) \]

5. **Substitute Values**:
\[ \text{Area} = \frac{1}{2} \times (10,000 + 50,000) \times \left(50 \times 10^{-6} - 200 \times 10^{-6}\right) \]
\[ = \frac{1}{2} \times 60,000 \times (-150 \times 10^{-6}) \]
\[ = \frac{1}{2} \times 60,000 \times -0.00015 \]
\[ = -4.5 \, \text{J} \]

6. **Interpret the Result**: The negative sign indicates that work is done on the system.

### Part (b): Change in Internal Energy of the Gas

1. **Use the First Law of Thermodynamics**: The first law states:
\[ \Delta U = Q - W \]
where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.

2. **Given Conditions**: Since no heat is supplied or extracted from the gas, \( Q = 0 \).

3. **Substitute Values**:
\[ \Delta U = 0 - (-4.5) = 4.5 \, \text{J} \]

4. **Conclusion**: The change in internal energy of the gas is \( 4.5 \, \text{J} \).

### Summary of Results
- (a) The work done by the gas is \( -4.5 \, \text{J} \).
- (b) The change in internal energy of the gas is \( 4.5 \, \text{J} \).