In a system, a piston caused an external pressure of `1.25` bar giving a change in volume of `32 L` for which, `Delta E=-51KJ`. What was the value of heat involved`:`

Options

A. `-55kJ`
B. `-11kJ`
C. `-47kJ`
D. `-91kJ`

Explanation

To find the value of heat involved in the given system, we can use the first law of thermodynamics, which states:

\[ \Delta E = Q + W \]

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

Given:
- \(\Delta E = -51 \, \text{kJ}\)
- External pressure \(P = 1.25 \, \text{bar}\)
- Change in volume \(\Delta V = 32 \, \text{L}\)

### Step 1: Calculate the work done (W)

The work done by the system can be calculated using the formula:

\[ W = -P \Delta V \]

First, we need to convert the pressure from bar to kJ/L. Since \(1 \, \text{bar} = 100 \, \text{kPa}\) and \(1 \, \text{kPa} = 0.001 \, \text{kJ/L}\), we have:

\[ 1.25 \, \text{bar} = 1.25 \times 100 \, \text{kPa} = 125 \, \text{kPa} = 0.125 \, \text{kJ/L} \]

Now we can calculate the work done:

\[ W = - (0.125 \, \text{kJ/L}) \times (32 \, \text{L}) = -4 \, \text{kJ} \]

### Step 2: Apply the first law of thermodynamics

Now we can substitute the values into the first law of thermodynamics equation:

\[ \Delta E = Q + W \]

Rearranging gives us:

\[ Q = \Delta E - W \]

Substituting the values we have:

\[ Q = -51 \, \text{kJ} - (-4 \, \text{kJ}) = -51 \, \text{kJ} + 4 \, \text{kJ} = -47 \, \text{kJ} \]

### Final Answer

The value of heat involved \(Q\) is:

\[ Q = -47 \, \text{kJ} \]