A diatomic gas, having `C_(p)=(7)/(2)R and C_(v)=(5)/(2)R` is heated at constant pressure. The ratio `dU : dQ : dW`

Options

A. `5:7:3`
B. `5:7:2`
C. `3:7:2`
D. `3:5:2`

Answer

To find the ratio \( dU : dQ : dW \) for a diatomic gas heated at constant pressure, we can follow these steps:

### Step 1: Understand the definitions
- \( dU \) is the change in internal energy.
- \( dQ \) is the heat added to the system.
- \( dW \) is the work done by the system.

### Step 2: Use the formulas for each quantity
1. **Change in internal energy**:
\[ dU = N C_v dT \]
where \( C_v = \frac{5}{2} R \) for a diatomic gas.

2. **Heat added at constant pressure**:
\[ dQ = N C_p dT \]
where \( C_p = \frac{7}{2} R \) for a diatomic gas.

3. **Work done at constant pressure**:
\[ dW = P dV \]
Using the ideal gas law, \( P dV = N R dT \).

### Step 3: Substitute the values
- For \( dU \):
\[ dU = N \left(\frac{5}{2} R\right) dT \]

- For \( dQ \):
\[ dQ = N \left(\frac{7}{2} R\right) dT \]

- For \( dW \):
\[ dW = N R dT \]

### Step 4: Write the ratios
Now we can write the ratios:
\[ dU : dQ : dW = N \left(\frac{5}{2} R\right) dT : N \left(\frac{7}{2} R\right) dT : N R dT \]

### Step 5: Simplify the ratios
We can cancel \( N \) and \( dT \) from all terms:
\[ \frac{5}{2} R : \frac{7}{2} R : R \]

This simplifies to:
\[ \frac{5}{2} : \frac{7}{2} : 1 \]

### Step 6: Multiply to eliminate fractions
To eliminate the fractions, multiply each term by 2:
\[ 5 : 7 : 2 \]

### Final Answer
Thus, the ratio \( dU : dQ : dW \) is:
\[ \boxed{5 : 7 : 2} \]