Consider one mole of helium gas enclosed in a container at initial pressure `P_(1)`and volume `V_(1)` . It expands isothermally to volume `V_(1)` After this , the gas expands adiabatically and its volume becomes `32V_(1)` . The work done by the gas during isothermal and adiabatic expansion processes are `W_(iso)andW_(adia)` respectively . if the radio `(W_(iso))/(W_(adia))=fIn 2`, then f is __________ .

Explanation

To solve the problem, we need to calculate the work done during the isothermal and adiabatic expansions of one mole of helium gas. Let's break it down step by step.

### Step 1: Work Done During Isothermal Expansion

For an isothermal process, the work done \( W_{iso} \) can be calculated using the formula:

\[ W_{iso} = nRT \ln\left(\frac{V_f}{V_i}\right) \]

Where:
- \( n \) = number of moles (1 mole for helium)
- \( R \) = universal gas constant
- \( T \) = absolute temperature
- \( V_f \) = final volume after isothermal expansion
- \( V_i \) = initial volume

Given that the gas expands isothermally from volume \( V_1 \) to \( 4V_1 \):

\[ W_{iso} = 1 \cdot R \cdot T \ln\left(\frac{4V_1}{V_1}\right) = RT \ln(4) \]

### Step 2: Work Done During Adiabatic Expansion

For an adiabatic process, the work done \( W_{adia} \) can be derived from the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

Since \( Q = 0 \) for an adiabatic process, we have:

\[ W_{adia} = -\Delta U \]

The change in internal energy \( \Delta U \) for an ideal gas is given by:

\[ \Delta U = nC_v \Delta T \]

Where \( C_v \) for a monoatomic gas (like helium) is:

\[ C_v = \frac{3R}{2} \]

Now, we need to determine the change in temperature \( \Delta T \). Using the adiabatic condition:

\[ TV^{\gamma - 1} = \text{constant} \]

Where \( \gamma = \frac{C_p}{C_v} = \frac{5}{3} \) for a monoatomic gas.

Using the initial and final volumes:
- Initial volume \( V_2 = 4V_1 \)
- Final volume \( V_f = 32V_1 \)

From the adiabatic condition:

\[ T_1 (4V_1)^{\frac{2}{3}} = T_2 (32V_1)^{\frac{2}{3}} \]

Solving for \( T_2 \):

\[ T_2 = T_1 \left(\frac{4}{32}\right)^{\frac{2}{3}} = T_1 \left(\frac{1}{8}\right)^{\frac{2}{3}} = T_1 \cdot \frac{1}{4} \]

Thus, the change in temperature is:

\[ \Delta T = T_2 - T_1 = \frac{T_1}{4} - T_1 = -\frac{3T_1}{4} \]

Now substituting into the work done formula:

\[ W_{adia} = -nC_v \Delta T = -1 \cdot \frac{3R}{2} \cdot \left(-\frac{3T_1}{4}\right) = \frac{9RT_1}{8} \]

### Step 3: Finding the Ratio \( \frac{W_{iso}}{W_{adia}} \)

Now we can find the ratio of the work done during the isothermal and adiabatic processes:

\[ \frac{W_{iso}}{W_{adia}} = \frac{RT \ln(4)}{\frac{9RT_1}{8}} = \frac{8 \ln(4)}{9} \]

Since \( \ln(4) = \ln(2^2) = 2 \ln(2) \):

\[ \frac{W_{iso}}{W_{adia}} = \frac{8 \cdot 2 \ln(2)}{9} = \frac{16 \ln(2)}{9} \]

### Conclusion

Given that \( \frac{W_{iso}}{W_{adia}} = f \ln(2) \), we can equate:

\[ f = \frac{16}{9} \]

Thus, the final answer is:

\[ f = \frac{16}{9} \]