a container with a specific volume \v\ changes to \2v.\ what happens to the average distance between gas…
a container with a specific volume \v\ changes to \2v.\ what happens to the average distance between gas molecules? assume that pressure and temperature of the gas remain constant.
Answer
Explanation:
Step1: Recall ideal - gas behavior
The number of moles of gas is constant as pressure ($P$), temperature ($T$) and amount of gas ($n$) are constant (from $PV = nRT$). When the volume doubles from $V$ to $2V$, the gas expands.
Step2: Consider the three - dimensional nature
The volume of a cube (a simple model for the space occupied by gas) is $V = L^3$, where $L$ is the side - length of the cube. If the initial volume is $V_1 = L_1^3$ and the final volume is $V_2 = L_2^3$, and $V_2 = 2V_1$. Then $L_2^3=2L_1^3$, so $L_2 = L_1\times2^{\frac{1}{3}}$. The average distance between gas molecules is proportional to the side - length of the cube representing the volume of the gas.
Answer:
The average distance between gas molecules increases by a factor of $2^{\frac{1}{3}}$.