Question

consider 1 mole of a real gas occupying a volume of v at a pressure p and temperature t. the ideal gas law would predict a pressure p_ideal. the actual pressure p_real is related to p_ideal by the van der waals equation. 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 $P_{ideal}$.
\[P_{ideal}=\frac{nRT}{V}\]
\[P_{ideal}=\frac{1.0\ mol\times0.0821\ L\cdot atm/mol\cdot K\times300\ K}{10.0\ L}\]
\[P_{ideal}= 2.463\ atm\]

Step2: Analyze real - gas behavior with van der Waals equation

The van der Waals equation for real gases is $(P + \frac{n^{2}a}{V^{2}})(V - nb)=nRT$. For a gas with non - zero intermolecular forces (like methane), the term $\frac{n^{2}a}{V^{2}}$ accounts for the intermolecular attractions. These attractions reduce the force of the gas molecules hitting the container walls, thus reducing the pressure. Since methane has non - zero intermolecular forces, $P_{real}

Answer:

For the first multiple - choice question: A. $P_{real}For the second part: The pressure calculated using the ideal gas law is $P_{ideal}=2.463\ atm$. The van der Waals equation would predict a lower pressure ($P_{real}