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. (a)1000 litres
B. (b)100 litres
C. (c)10 litres
D. (d)1 litre

Answer

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

### Step 1: Convert heat energy from calories to joules
The heat energy supplied to the gas is given as 143 calories. To convert this to joules, we use the conversion factor:
\[ 1 \text{ calorie} = 4.2 \text{ joules} \]
So, we calculate:
\[ \Delta Q = 143 \text{ calories} \times 4.2 \text{ joules/calorie} = 600.6 \text{ joules} \]

### Step 2: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ \Delta Q = \Delta U + W \]
Where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done by the system.

Given that \(\Delta U = 500 \text{ joules}\), we can rearrange the equation to find the work done:
\[ W = \Delta Q - \Delta U = 600.6 \text{ joules} - 500 \text{ joules} = 100.6 \text{ joules} \]

### Step 3: Relate work done to change in volume
At constant pressure, the work done by the gas can be expressed as:
\[ W = P \Delta V \]
Where:
- \(P\) is the pressure,
- \(\Delta V\) is the change in volume.

We know that the pressure \(P = 10^5 \text{ N/m}^2\). We can rearrange the equation to solve for \(\Delta V\):
\[ \Delta V = \frac{W}{P} \]

### Step 4: Substitute the values to find the change in volume
Now we substitute the values we have:
\[ \Delta V = \frac{100.6 \text{ joules}}{10^5 \text{ N/m}^2} \]
Calculating this gives:
\[ \Delta V = \frac{100.6}{10^5} \text{ m}^3 = 1.006 \times 10^{-3} \text{ m}^3 \]

### Step 5: Convert the change in volume to liters
Since \(1 \text{ m}^3 = 1000 \text{ liters}\), we convert:
\[ \Delta V = 1.006 \times 10^{-3} \text{ m}^3 \times 1000 \text{ liters/m}^3 = 1.006 \text{ liters} \]
Rounding this, we find:
\[ \Delta V \approx 1 \text{ liter} \]

### Final Answer
Thus, the change in volume of the gas is approximately:
\[ \Delta V = 1 \text{ liter} \]