One mole of a monatomic gas at pressure 2 atm ,279 K taken to final pressure 4 atm by a reversible path described by `P//V` = constant . Calculate the magnitude of `(DeltaE)/(w)` for the process .

Answer

To solve the problem, we need to calculate the magnitude of \(\frac{\Delta E}{W}\) for a process involving a monatomic gas. Let's break down the steps:

### Step 1: Identify the Initial Conditions
We have:
- Initial pressure, \(P_1 = 2 \, \text{atm}\)
- Initial temperature, \(T_1 = 279 \, \text{K}\)
- Final pressure, \(P_2 = 4 \, \text{atm}\)
- Number of moles, \(n = 1\)

### Step 2: Determine the Final Temperature
Since the process follows the path described by \(P/V = \text{constant}\), we can use the ideal gas law to find the final temperature \(T_2\).

Using the ideal gas equation:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Assuming \(P = k/V\), we can express \(V_1\) and \(V_2\) in terms of \(k\):
\[ V_1 = \frac{nRT_1}{P_1} \quad \text{and} \quad V_2 = \frac{nRT_2}{P_2} \]

Substituting \(V_1\) and \(V_2\) into the equation gives:
\[ \frac{P_1 \cdot \frac{nRT_1}{P_1}}{T_1} = \frac{P_2 \cdot \frac{nRT_2}{P_2}}{T_2} \]

This simplifies to:
\[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]

Substituting \(V_1\) and \(V_2\):
\[ T_2 = \frac{4 \cdot \frac{nRT_2}{4}}{2 \cdot \frac{nRT_1}{2}} \cdot T_1 \]

This leads to:
\[ T_2 = 4T_1 \]

### Step 3: Calculate \(\Delta E\)
The change in internal energy for a monatomic gas is given by:
\[ \Delta E = n C_V \Delta T \]
Where \(C_V\) for a monatomic gas is \(\frac{3}{2}R\).

Calculating \(\Delta T\):
\[ \Delta T = T_2 - T_1 = 4T_1 - T_1 = 3T_1 \]

Now substituting into the equation for \(\Delta E\):
\[ \Delta E = 1 \cdot \frac{3}{2}R \cdot 3T_1 = \frac{9}{2}RT_1 \]

### Step 4: Calculate Work Done \(W\)
For a process where \(P/V = \text{constant}\), the work done can be calculated as:
\[ W = -\int_{V_1}^{V_2} P \, dV \]

Using \(P = k/V\), we have:
\[ W = -k \int_{V_1}^{V_2} \frac{1}{V} \, dV = -k \left[ \ln V \right]_{V_1}^{V_2} = -k (\ln V_2 - \ln V_1) = -k \ln \frac{V_2}{V_1} \]

Substituting \(V_1\) and \(V_2\):
\[ W = -\frac{nRT_1}{2} \ln \frac{4}{2} = -\frac{nRT_1}{2} \ln 2 \]

### Step 5: Calculate \(\frac{\Delta E}{W}\)
Now we can find \(\frac{\Delta E}{W}\):
\[ \frac{\Delta E}{W} = \frac{\frac{9}{2}RT_1}{-\frac{nRT_1}{2} \ln 2} = -\frac{9}{n \ln 2} \]

Substituting \(n = 1\):
\[ \frac{\Delta E}{W} = -\frac{9}{\ln 2} \]

### Step 6: Calculate Magnitude
The magnitude of \(\frac{\Delta E}{W}\) is:
\[ \left| \frac{\Delta E}{W} \right| = \frac{9}{\ln 2} \]

### Final Answer
The magnitude of \(\frac{\Delta E}{W}\) for the process is \(\frac{9}{\ln 2}\).

---