143 calories of heat energy are supplied to a gas . Its internal energy rises by 500 joules . If the pressure remains constant and is equal to `10^5 N m^(-2)` , the change in the volume of the gas is

Options

A. 1000 litres
B. 100 litres
C. 10 litres
D. 1 litre

Answer

To solve the problem, we will use the first law of thermodynamics and the relationship between work done and change in volume at constant pressure.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Heat energy supplied, \( Q = 143 \) calories
- Change in internal energy, \( \Delta U = 500 \) joules
- Constant pressure, \( P = 10^5 \) N/m²

2. **Convert Heat Energy from Calories to Joules**:
- We know that \( 1 \) calorie = \( 4.2 \) joules.
- Therefore, \( Q = 143 \times 4.2 = 600.6 \) joules.

3. **Apply the First Law of Thermodynamics**:
- According to the first law, \( Q = \Delta U + W \)
- Rearranging gives us \( W = Q - \Delta U \).
- Substituting the values we have:
\[ W = 600.6 \, \text{joules} - 500 \, \text{joules} = 100.6 \, \text{joules} \]

4. **Relate Work Done to Change in Volume**:
- At constant pressure, the work done on the gas can be expressed as:
\[ W = P \Delta V \]
- Rearranging gives us:
\[ \Delta V = \frac{W}{P} \]

5. **Substitute the Values into the Volume Change Equation**:
- Substituting \( W = 100.6 \, \text{joules} \) and \( P = 10^5 \, \text{N/m}^2 \):
\[ \Delta V = \frac{100.6}{10^5} = 1.006 \times 10^{-3} \, \text{m}^3 \]

6. **Convert Volume Change to Liters**:
- Since \( 1 \, \text{m}^3 = 1000 \, \text{liters} \):
\[ \Delta V = 1.006 \times 10^{-3} \, \text{m}^3 = 1.006 \, \text{liters} \]

### Final Answer:
The change in volume of the gas is approximately \( 1.006 \) liters.

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