A sample of an ideal gas is expanded from `1 dm^(3)` to 3 in a reversible process for which `P=KV^(3)`, with `K=1//5 (atm//dm^(3))`, what is work done by gas `(L atm)`

Answer

To calculate the work done by the gas during the reversible expansion, we can use the formula for work done in a reversible process:

\[ W = -\int_{V_1}^{V_2} P \, dV \]

Given:
- Initial volume, \( V_1 = 1 \, \text{dm}^3 \)
- Final volume, \( V_2 = 3 \, \text{dm}^3 \)
- Pressure, \( P = K V^3 \) where \( K = \frac{1}{5} \, \text{atm/dm}^3 \)

### Step 1: Substitute the expression for pressure into the work formula
We can substitute \( P \) into the work formula:

\[ W = -\int_{1}^{3} K V^3 \, dV \]

### Step 2: Substitute the value of \( K \)
Substituting \( K = \frac{1}{5} \):

\[ W = -\int_{1}^{3} \frac{1}{5} V^3 \, dV \]

### Step 3: Factor out the constant
We can factor out the constant \( \frac{1}{5} \):

\[ W = -\frac{1}{5} \int_{1}^{3} V^3 \, dV \]

### Step 4: Integrate \( V^3 \)
Now we integrate \( V^3 \):

\[ \int V^3 \, dV = \frac{V^4}{4} + C \]

### Step 5: Evaluate the definite integral
Now we evaluate the definite integral from 1 to 3:

\[ W = -\frac{1}{5} \left[ \frac{V^4}{4} \right]_{1}^{3} \]

Calculating the limits:

\[ W = -\frac{1}{5} \left( \frac{3^4}{4} - \frac{1^4}{4} \right) \]

\[ = -\frac{1}{5} \left( \frac{81}{4} - \frac{1}{4} \right) \]

\[ = -\frac{1}{5} \left( \frac{80}{4} \right) \]

\[ = -\frac{1}{5} \times 20 \]

### Step 6: Calculate the final value
\[ W = -4 \, \text{L atm} \]

### Final Answer
The work done by the gas during the expansion is \( W = -4 \, \text{L atm} \).

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