Question

One mole of an ideal gas is contained in a 1 L vessel at T(K) with a pressure of p atm. The vessel is split into two equal parts. What are the pressure and volume in each part?

• A. $p { atm}, \frac{1}{2} {L}$
• B. $p { atm}, 1 {L}$
• C. $2p { atm}, \frac{1}{2} {L}$
• D. $\frac{p}{2} { atm}, \frac{1}{2} {L}$

Hint:

When an ideal gas vessel is divided, use $PV = nRT$ to find the new pressure. Adjust for the number of moles and volume in each part while keeping temperature constant.

Answer 1

The Correct Option is A

Initially, 1 mol of gas occupies 1 L at pressure $p$ and temperature $T$.
Using the ideal gas law, $PV = nRT$:
$p \times 1 = 1 \times RT \implies p = RT$.
The vessel is divided into two equal parts, so each part has a volume of $\frac{1}{2}$ L, and the gas splits equally (0.5 mol per part).
For each part, apply the ideal gas law:
$P \times \frac{1}{2} = 0.5 \times RT \implies P \times \frac{1}{2} = 0.5 \times p \implies P = p$.
Thus, each part has a pressure of $p$ atm and a volume of $\frac{1}{2}$ L.