A volume of `10m^(3)` of a liquid is supplied with `100 kal` of heat and expands at a constant pressure of `10 atm` to a final volume of `10.2m^(3)`. Calculate the work done and change in internal energy.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Process
We have a liquid expanding at constant pressure, which indicates that this is an isobaric process.

### Step 2: Calculate the Work Done
The work done \( W \) during an isobaric process can be calculated using the formula:
\[ W = P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

Given:
- Initial volume \( V_1 = 10 \, m^3 \)
- Final volume \( V_2 = 10.2 \, m^3 \)
- Pressure \( P = 10 \, atm \)

First, calculate the change in volume:
\[ \Delta V = V_2 - V_1 = 10.2 \, m^3 - 10 \, m^3 = 0.2 \, m^3 \]

Now, convert the pressure from atm to pascals:
\[ 1 \, atm = 1.013 \times 10^5 \, Pa \quad \Rightarrow \quad P = 10 \, atm = 10 \times 1.013 \times 10^5 \, Pa = 1.013 \times 10^6 \, Pa \]

Now, substituting the values into the work formula:
\[ W = P \Delta V = (1.013 \times 10^6 \, Pa) \times (0.2 \, m^3) = 0.2026 \times 10^6 \, J = 202600 \, J \]

### Step 3: Convert Work Done to Kilocalories
To convert joules to kilocalories, we use the conversion factor:
\[ 1 \, kcal = 4186 \, J \]
Thus,
\[ W = \frac{202600 \, J}{4186 \, J/kcal} \approx 48.4 \, kcal \]

### Step 4: Calculate Change in Internal Energy
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
where \( Q \) is the heat added, \( \Delta U \) is the change in internal energy, and \( W \) is the work done.

Given:
- Heat supplied \( Q = 100 \, kcal \)
- Work done \( W \approx 48.4 \, kcal \)

Now, rearranging the equation to find \( \Delta U \):
\[ \Delta U = Q - W = 100 \, kcal - 48.4 \, kcal = 51.6 \, kcal \]

### Final Answers
- Work Done \( W \approx 48.4 \, kcal \)
- Change in Internal Energy \( \Delta U \approx 51.6 \, kcal \)

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