The value of `PV` for `5.6 L` of an ideal gas is ……… `RT` at `NTP`.

Explanation

To solve the question, we need to find the value of \( PV \) for \( 5.6 \, \text{L} \) of an ideal gas at Normal Temperature and Pressure (NTP), expressed in terms of \( RT \).

### Step-by-Step Solution:

1. **Understand the Ideal Gas Equation**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature

2. **Identify the Conditions at NTP**:
At Normal Temperature and Pressure (NTP):
- Temperature, \( T = 273.15 \, \text{K} \)
- Pressure, \( P = 1 \, \text{atm} \)
- Volume, \( V = 5.6 \, \text{L} \)

3. **Calculate the Number of Moles (\( n \))**:
Rearranging the ideal gas equation to find \( n \):
\[ n = \frac{PV}{RT} \]
Now, substituting the known values:
\[ n = \frac{(1 \, \text{atm}) \times (5.6 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \times (273.15 \, \text{K})} \]

4. **Perform the Calculation**:
Calculate the denominator:
\[ R \times T = 0.0821 \times 273.15 \approx 22.414 \, \text{L atm/mol} \]
Now, substituting back into the equation for \( n \):
\[ n = \frac{5.6}{22.414} \approx 0.2497 \, \text{mol} \approx 0.25 \, \text{mol} \]

5. **Express \( PV \) in terms of \( RT \)**:
Since we have calculated \( n \):
\[ PV = nRT = 0.25 \times RT \]

Thus, the value of \( PV \) for \( 5.6 \, \text{L} \) of an ideal gas at NTP is:
\[ PV = 0.25 \, RT \]

### Final Answer:
The value of \( PV \) for \( 5.6 \, \text{L} \) of an ideal gas is \( 0.25 \, RT \).

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