Four moles of an ideal diatomic gas `(gamma = 1.4)` at `300 K` and `12 atm` expanded irreversibly & adiabatically to a final pressure of `2.4 atm` against a constant pressure of `2 atm`. Calcualate `W` & `DeltaH` (magnitude only) . Report your answer as `(|X + Y|)/(10)`, where `W = X xx 10R, DeltaH = Y xx 10R` and `R` is gas constant.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify Given Values
- Number of moles (n) = 4 moles
- Initial temperature (T1) = 300 K
- Initial pressure (P1) = 12 atm
- Final pressure (P2) = 2.4 atm
- External pressure (P_ext) = 2 atm
- Heat capacity ratio (γ) = 1.4

### Step 2: Calculate the Change in Internal Energy (ΔU)
For an adiabatic process, the change in internal energy (ΔU) is equal to the work done (W) on the system.

Using the formula for ΔU in terms of moles and temperature change:
\[ \Delta U = n C_v (T_2 - T_1) \]
Where \( C_v \) for a diatomic gas is given by:
\[ C_v = \frac{5}{2} R \]
Thus,
\[ \Delta U = n \left(\frac{5}{2} R\right) (T_2 - T_1) \]

### Step 3: Use the Ideal Gas Law to Find T2
Using the ideal gas law and the relationship between pressures and volumes, we can find the final temperature (T2).

From the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Since the process is irreversible and adiabatic, we can also use the relation:
\[ \frac{T_2}{T_1} = \frac{P_2}{P_1} \left(\frac{V_1}{V_2}\right) \]

However, we can also calculate the work done against the external pressure:
\[ W = -P_{ext} (V_2 - V_1) \]

### Step 4: Calculate Work Done (W)
The work done in an irreversible expansion against a constant external pressure is given by:
\[ W = -P_{ext} \Delta V \]
Where \( \Delta V = V_2 - V_1 \).

Using the ideal gas law to express volumes:
\[ V = \frac{nRT}{P} \]
We can express \( V_1 \) and \( V_2 \):
\[ V_1 = \frac{nRT_1}{P_1} = \frac{4R \cdot 300}{12} = 100R \]
\[ V_2 = \frac{nRT_2}{P_2} \]

### Step 5: Calculate T2
Using the adiabatic condition and the pressures:
\[ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}} = \left(\frac{2.4}{12}\right)^{\frac{0.4}{1.4}} \]
Calculating this gives:
\[ T_2 = 300 \times \left(\frac{2.4}{12}\right)^{\frac{0.4}{1.4}} \approx 240 K \]

### Step 6: Substitute T2 back to find W
Now substituting \( T_2 \) back into the work done equation:
\[ W = -P_{ext} \left(\frac{nRT_2}{P_2} - \frac{nRT_1}{P_1}\right) \]
Calculating this gives:
\[ W = -2 \cdot \left(\frac{4R \cdot 240}{2.4} - 100R\right) = -2 \cdot (400R - 100R) = -600R \]

### Step 7: Calculate ΔH
The change in enthalpy (ΔH) is given by:
\[ \Delta H = n C_p (T_2 - T_1) \]
Where \( C_p = C_v + R = \frac{7}{2} R \):
\[ \Delta H = n \left(\frac{7}{2} R\right) (T_2 - T_1) = 4 \cdot \frac{7}{2} R \cdot (240 - 300) = 4 \cdot \frac{7}{2} R \cdot (-60) = -840R \]

### Step 8: Final Calculation
Now we have:
- \( W = -600R \) → \( X = -60 \)
- \( \Delta H = -840R \) → \( Y = -84 \)

Thus, we need to calculate:
\[ \frac{|X + Y|}{10} = \frac{|-60 - 84|}{10} = \frac{144}{10} = 14.4 \]

### Final Answer
The final answer is:
\[ \boxed{14.4} \]