Question

at a certain p and t, a gas occupies 20 l. if p and t are held constant, what will be the volume if half the gas sample escapes? a. 2 l b. 10 l c. 20 l d. 40 l \\(\frac{v_1}{n_1} = \frac{v_2}{n_2}\\)

Explanation:

Step1: Identify the gas law

This problem uses Avogadro's Law, which states that at constant pressure (\(P\)) and temperature (\(T\)), the volume (\(V\)) of a gas is directly proportional to the amount of gas (moles, \(n\)): \(\frac{V_1}{n_1}=\frac{V_2}{n_2}\).

Step2: Define initial and final conditions

Let the initial volume \(V_1 = 20\space L\) and initial moles \(n_1 = n\). After half the gas escapes, the final moles \(n_2=\frac{n}{2}\) (since half the sample is gone, the amount of gas is halved).

Step3: Apply Avogadro's Law

Substitute into the formula: \(\frac{20}{n}=\frac{V_2}{\frac{n}{2}}\).

Step4: Solve for \(V_2\)

Cross - multiply: \(n\times V_2 = 20\times\frac{n}{2}\).
Simplify the right - hand side: \(20\times\frac{n}{2}=10n\).
Then, divide both sides by \(n\) (assuming \(n
eq0\), which is valid for a gas sample): \(V_2 = 10\space L\).

Answer:

B. 10 L