The physical state of a gaseous substance is specified by giving its volume, pressure, temperature and chemical amount. There are two equations interrelating these four variables, namely the perfect gas equation and the van der Waals equation. 2.0 mol C2H6 is confined at 1000K in 200 cm3. What is the pressure (in atm) of the gas under this condition when it behaves as a perfect gas? R = 8.314 J/molK = 0.08206 L atm/molK

Question

BlackTom AI · Accepted Answer

We’re given a gaseous system with 2.0 mol of C2H6 confined at 1000 K in a volume of 200 cm3, and we are asked to find the pressure assuming ideal gas behavior.

First, convert all quantities to the appropriate units for the ideal gas equation PV = nRT. The volume is 200 cm3, which equals 0.200 L (since 1 L = 1000 cm3). The temperature is already given in kelvin as 1000 K. The number of moles is n = 2.0 mol.

Next, identify the appropriate constant R for units L, atm, mol, and K. Here R = 0.08206 L atm / (mol K).

Now apply the ideal gas law to solve for P: P = nRT / V.

Plugging in the values: P = (2.0 mol) × (0.08206 L atm / (mol K)) × (1000 K) ÷ (0.200 L).

Calculate step by step: 2.0 × 0.08206 = 0.16412; then 0.16412 × 1000 = 164.12; finally 164.12 ÷ 0.200 = 820.6.

Analyze the units to ensure consistency: mol cancels, L cancels, K cancels appropriately, leaving atm for pressure, as expected.

Consider potential pitfalls: ensuring the volume is in liters and temperature in kelvin, and using the correct R value corresponding to those units; mixing unit systems (e.g., using R in J/(mol K) with liters and atm) would yield incorrect results. If the gas did not behave ideally, the van der Waals corrections would be needed, but the problem specifies ideal gas behavior, so the ideal gas law provides the correct pressure here.

In summary, applying P = nRT / V with the given quantities yields P = 820.6 atm, confirming the supplied numeric result.

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