Question

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 \(^{-1}(R=0.082\) litre atm \(\left.\mathrm{mol}^{-1} \mathrm{deg}^{-1}\right)\) (1) \(\Lambda\) t \(\mathrm{STP}\) (2) When \(V=22.4\) litrc (3) When \(T=12 \mathrm{~K}\) (4) Impossible under any condition

Short answer

The conditions under which the pure sample of the ideal gas will exhibit the stated properties is at approximately T = 12 K.

Step-by-step solution

Understand the Ideal Gas Law

The Ideal Gas Law is given by: \[ PV = nRT \]Where: \(P\) = Pressure\(V\) = Volume\(n\) = Number of moles\(R\) = Universal gas constant \(T\) = Temperature

Given Conditions

The problem states that the pressure \(P\) is 1 atm and the concentration \(C\) is 1 mol/L. Concentration \(C\) is defined as the number of moles \(n\) per unit volume \(V\), or \(C = \frac{n}{V}\). With 1 mol/litre concentration, \(V = 1\) litre and \(n = 1\) mol.

Substitute Values into the Ideal Gas Law

Substitute \(P = 1\, \text{atm}\), \(n = 1\, \text{mol}\), \(V = 1\, \text{litre}\), and \(R = 0.082\, \text{litre}\, \text{atm}\, \text{mol}^{-1}\, \text{deg}^{-1}\) into the Ideal Gas Law: \[ 1 \cdot 1 = 1 \cdot 0.082 \cdot T \]

Solve for Temperature \(T\)

Rearrange the equation to solve for \(T\):\[ T = \frac{1 \cdot 1}{1 \cdot 0.082} = \frac{1}{0.082} = 12.2 \text{ K} \]

Compare to Given Options

The calculated temperature (12.2 K) is very close to one of the given conditions (12 K). Though not identical, it is reasonably close for practical purposes.

Key concepts

Pressure

Pressure is a key concept in the Ideal Gas Law. It is the force exerted by gas particles per unit area on the walls of the container they are in. For gases, pressure can be measured in various units, with atmospheres (atm) being a common one for ideal gas problems. In the Ideal Gas Law equation \(PV = nRT\), pressure (\