Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of `1 atm` but also a concentration of `1 mol litre^(-1)`
`[R= 0.082` iltre atm `mol^(-1)K^(-1)]`

Options

A. At STP
B. When V = 22.4 L
C. When T = 12 K
D. Impossible under any conditions

Explanation

To determine the conditions under which a pure sample of an ideal gas exhibits a pressure of 1 atm and a concentration of 1 mol/L, we can use the ideal gas equation:

### Step-by-Step Solution:

1. **Write the Ideal Gas Equation**:
The ideal gas equation is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (0.082 L atm K\(^{-1}\) mol\(^{-1}\))
- \( T \) = temperature (in Kelvin)

2. **Convert to Concentration**:
We know that concentration \( C \) is defined as:
\[ C = \frac{n}{V} \]
Rearranging the ideal gas equation, we can express pressure in terms of concentration:
\[ P = CRT \]

3. **Substitute Known Values**:
Given:
- \( P = 1 \) atm
- \( C = 1 \) mol/L
- \( R = 0.082 \) L atm K\(^{-1}\) mol\(^{-1}\)

Substitute these values into the equation:
\[ 1 = (1)(0.082)T \]

4. **Solve for Temperature \( T \)**:
Rearranging the equation gives:
\[ T = \frac{1}{0.082} \]
Calculating this:
\[ T \approx 12.2 \text{ K} \]

5. **Conclusion**:
The conditions under which a pure sample of an ideal gas exhibits a pressure of 1 atm and a concentration of 1 mol/L is at a temperature of approximately 12.2 K.

### Final Answer:
- The ideal gas will exhibit a pressure of 1 atm and a concentration of 1 mol/L at a temperature of approximately 12.2 K.