Question

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: Calculate pressure using ideal gas law

The ideal gas law is $PV = nRT$. We need to solve for $P$, so $P=\frac{nRT}{V}$. Given $n = 1.0\ mol$, $R=0.0821\ L\cdot atm/mol\cdot K$, $T = 300\ K$ and $V=10.0\ L$. Substitute the values: $P=\frac{1.0\ mol\times0.0821\ L\cdot atm/mol\cdot K\times300\ K}{10.0\ L}=2.463\ atm$.

Step2: Analyze van der Waals correction

The van der Waals equation is $(P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT$. For real - gases with non - zero intermolecular forces (represented by the $a$ term) and non - zero molecular volume (represented by the $b$ term), the correction due to intermolecular forces is significant here. Methane has weak intermolecular forces. The term $\frac{n^{2}a}{V^{2}}$ is positive. When we solve the van der Waals equation for $P$ we get $P=\frac{nRT}{V - nb}-\frac{n^{2}a}{V^{2}}$. The first term $\frac{nRT}{V - nb}$ would be larger than $\frac{nRT}{V}$ (from the ideal gas law) if we only consider the volume correction, but the second term $\frac{n^{2}a}{V^{2}}$ subtracts from it. Since the intermolecular forces are non - zero, the real pressure $P_{real}$ is less than the ideal pressure $P_{ideal}$ because the intermolecular forces tend to pull the gas molecules together and reduce the force of their impact on the container walls.

Answer:

A. $P_{real}