The number of moles of an ideal gas that should be taken in a closed vessel of 30L capacity at a temperature of `27^(@)C` so that the pressure exerted by the gas on the walls of the container is 4.1 atmosphere is

Answer

To solve the problem of finding the number of moles of an ideal gas in a closed vessel, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atmospheres)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert the temperature from Celsius to Kelvin:**
\[ T(K) = T(°C) + 273.15 \]
Given \( T = 27°C \):
\[ T = 27 + 273.15 = 300.15 \, K \]

2. **Identify the values for the variables:**
- Pressure, \( P = 4.1 \, \text{atm} \)
- Volume, \( V = 30 \, \text{L} \)
- Ideal gas constant, \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- Temperature, \( T = 300.15 \, K \)

3. **Rearrange the Ideal Gas Law to solve for \( n \):**
\[ n = \frac{PV}{RT} \]

4. **Substitute the values into the equation:**
\[ n = \frac{(4.1 \, \text{atm}) \times (30 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)}) \times (300.15 \, K)} \]

5. **Calculate the numerator:**
\[ 4.1 \times 30 = 123 \, \text{atm·L} \]

6. **Calculate the denominator:**
\[ 0.0821 \times 300.15 \approx 24.726 \, \text{L·atm/(K·mol)} \]

7. **Calculate the number of moles \( n \):**
\[ n = \frac{123}{24.726} \approx 4.97 \, \text{moles} \]

8. **Round the answer to two decimal places:**
\[ n \approx 5.00 \, \text{moles} \]

### Final Answer:
The number of moles of the ideal gas required is approximately **5.00 moles**.