A gas under constant pressure of `4.5 xx 10^(5) Pa` when subjected to `800 kJ` of heat, changes the volume from `0.5 m^(3) to 2.0 m^(3)`. The change in internal energy of the gas is

Options

A. `6.75xx10^5 J`
B. `5.25xx10^5 J`
C. `3.25xx10^5 J`
D. `1.25xx10^5 J`

Answer

To find the change in internal energy of the gas, we can use the first law of thermodynamics, which states:

\[ Q = \Delta U + W \]

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

### Step 1: Identify the given values
- Heat added, \( Q = 800 \, \text{kJ} = 800 \times 10^3 \, \text{J} \)
- Constant pressure, \( P = 4.5 \times 10^5 \, \text{Pa} \)
- Initial volume, \( V_i = 0.5 \, \text{m}^3 \)
- Final volume, \( V_f = 2.0 \, \text{m}^3 \)

### Step 2: Calculate the work done by the gas
At constant pressure, the work done \( W \) is given by:

\[ W = P \Delta V \]

Where \( \Delta V \) is the change in volume:

\[ \Delta V = V_f - V_i = 2.0 \, \text{m}^3 - 0.5 \, \text{m}^3 = 1.5 \, \text{m}^3 \]

Now substitute the values into the work done equation:

\[ W = P \Delta V = (4.5 \times 10^5 \, \text{Pa}) \times (1.5 \, \text{m}^3) \]

Calculating this gives:

\[ W = 4.5 \times 10^5 \times 1.5 = 6.75 \times 10^5 \, \text{J} = 675 \, \text{kJ} \]

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

\[ \Delta U = Q - W \]

Substituting the known values:

\[ \Delta U = 800 \times 10^3 \, \text{J} - 675 \times 10^3 \, \text{J} \]

Calculating this gives:

\[ \Delta U = 800000 - 675000 = 125000 \, \text{J} \]

### Step 4: Convert the result to kilojoules
\[ \Delta U = 125000 \, \text{J} = 125 \, \text{kJ} \]

### Final Answer
The change in internal energy of the gas is:

\[ \Delta U = 125 \, \text{kJ} \]