Question

for a gas with significant intermolecular attractions, how would p_real compare to p_ideal?
p_real < p_ideal
p_real = p_ideal
p_real > p_ideal
the relationship cannot be determined without knowing the values of a and b
calculate the approximate pressure exerted by 1.0 mole of methane (ch₄) in a 10.0 l container at 300 k using the ideal gas law (r = 0.0821 l·atm/mol·k). then, qualitatively explain how the van der waals equation would likely modify this pressure, considering that methane has weak but non - zero intermolecular forces and a small molecular volume.

Explanation:

Step1: Recall ideal gas law

The ideal gas law is $PV = nRT$. We are given $n = 1.0\ mol$, $V=10.0\ L$, $T = 300\ K$ and $R=0.0821\ L\cdot atm/mol\cdot K$. We can solve for $P$ (ideal pressure).
$P=\frac{nRT}{V}$

Step2: Calculate ideal pressure

Substitute the values into the formula:
$P=\frac{1.0\ mol\times0.0821\ L\cdot atm/mol\cdot K\times300\ K}{10.0\ L}= 2.463\ atm$

Step3: Analyze real - gas behavior

The van der Waals equation is $(P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT$. For a gas with intermolecular forces (like methane), the term $\frac{n^{2}a}{V^{2}}$ accounts for the attractive forces between molecules. These attractive forces reduce the force of the gas molecules hitting the container walls, so the real pressure $P_{real}$ is less than the ideal pressure $P_{ideal}$. The $nb$ term accounts for the volume of the molecules themselves, but in this case, the main effect is due to the intermolecular attractions.

Answer:

The first part: The pressure calculated using the ideal - gas law is $2.463\ atm$.
The second part: The real pressure $P_{real}$ of methane is less than the ideal pressure $P_{ideal}$ because the intermolecular forces in methane reduce the force of the gas molecules hitting the container walls. The answer to the multiple - choice part is $P_{real}