For the reaction
`2A(g)+3B(l)rarrC(g)+4D(l),DeltaH=300"cal//mol"`
Calculate `DeltaU` (in calories) of reaction when `3`moles of `A` (g) react with `4` moles of B (l) at `27^(@)C.` `(R=2"cal//mol"-K)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Reaction and Given Data
The reaction is:
\[ 2A(g) + 3B(l) \rightarrow C(g) + 4D(l) \]
We are given:
- \(\Delta H = 300 \, \text{cal/mol}\)
- 3 moles of \(A\) and 4 moles of \(B\) are reacting.
- Temperature \(T = 27^\circ C = 300 \, K\) (since \(27 + 273 = 300\))
- \(R = 2 \, \text{cal/mol-K}\)

### Step 2: Determine the Limiting Reagent
From the stoichiometry of the reaction:
- 2 moles of \(A\) react with 3 moles of \(B\).
- Therefore, 1 mole of \(B\) reacts with \(\frac{2}{3}\) moles of \(A\).

For 4 moles of \(B\):
- Moles of \(A\) required = \(4 \times \frac{2}{3} = \frac{8}{3}\) moles.

Since we only have 3 moles of \(A\), \(A\) is the limiting reagent.

### Step 3: Calculate the Amount of Reactants and Products
Using the limiting reagent \(A\):
- Moles of \(B\) that will react with 3 moles of \(A\):
\[ \text{Moles of } B \text{ required} = 3 \times \frac{3}{2} = 4.5 \text{ moles} \]
- Since we only have 4 moles of \(B\), we will use all 4 moles of \(B\).

### Step 4: Calculate the Moles of Products Formed
From the stoichiometry:
- For 2 moles of \(A\), 1 mole of \(C\) is produced.
- Therefore, for 3 moles of \(A\):
\[ \text{Moles of } C = \frac{3}{2} = 1.5 \text{ moles} \]
- Moles of \(D\) produced:
\[ \text{Moles of } D = \frac{4}{2} \times 3 = 6 \text{ moles} \]

### Step 5: Calculate \(\Delta N_G\)
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 2 = -1 \]

### Step 6: Use the Relationship Between \(\Delta H\) and \(\Delta U\)
The relationship is given by:
\[ \Delta H = \Delta U + \Delta N_G RT \]
Substituting the known values:
\[ 300 = \Delta U + (-1)(2)(300) \]
\[ 300 = \Delta U - 600 \]
\[ \Delta U = 300 + 600 = 900 \, \text{calories} \]

### Step 7: Adjust for the Amount of Reaction
Since we are considering the reaction of 3 moles of \(A\) with 4 moles of \(B\), we need to scale \(\Delta U\) accordingly:
\[ \Delta U \text{ for 3 moles of } A = 900 \times \frac{3}{2} = 1350 \, \text{calories} \]

### Final Answer
Thus, the change in internal energy \(\Delta U\) for the reaction when 3 moles of \(A\) react with 4 moles of \(B\) is:
\[ \Delta U = 1350 \, \text{calories} \]