. When heat energy of 1500 Joules , is supplied to a gas at constant pressure `2.1xx10^(5)N//m^(2),` there was an increase in its volume equal to `2.5xx10^(-3)m^(3)`.The increase in internal energy of the gas in joules is

Options

A. 45
B. 525
C. 975
D. Zero

Answer

To find the increase in internal energy of the gas, we can use the first law of thermodynamics, which is given by the equation:

\[ \Delta Q = \Delta U + \Delta W \]

Where:
- \(\Delta Q\) is the heat supplied to the system (in joules),
- \(\Delta U\) is the increase in internal energy (in joules),
- \(\Delta W\) is the work done by the system (in joules).

### Step 1: Identify the given values
From the problem statement, we have:
- Heat energy supplied, \(\Delta Q = 1500 \, \text{J}\)
- Constant pressure, \(P = 2.1 \times 10^5 \, \text{N/m}^2\)
- Increase in volume, \(\Delta V = 2.5 \times 10^{-3} \, \text{m}^3\)

### Step 2: Calculate the work done by the gas
At constant pressure, the work done by the gas can be calculated using the formula:

\[ \Delta W = P \Delta V \]

Substituting the values:

\[ \Delta W = (2.1 \times 10^5 \, \text{N/m}^2) \times (2.5 \times 10^{-3} \, \text{m}^3) \]

Calculating this:

\[ \Delta W = 2.1 \times 2.5 \times 10^5 \times 10^{-3} = 2.1 \times 2.5 \times 10^2 = 5.25 \times 10^2 = 525 \, \text{J} \]

### Step 3: Use the first law of thermodynamics to find \(\Delta U\)
Now, we can rearrange the first law of thermodynamics to find the increase in internal energy:

\[ \Delta U = \Delta Q - \Delta W \]

Substituting the values we have:

\[ \Delta U = 1500 \, \text{J} - 525 \, \text{J} \]

Calculating this gives:

\[ \Delta U = 975 \, \text{J} \]

### Conclusion
The increase in internal energy of the gas is:

\[ \Delta U = 975 \, \text{J} \]