The molar volume of oxygen, in L, at 1.00 atmosphere and 100°C, is closest to A - VCE - SSCE Chemistry - Question 2 - 2006 - Paper 1

To find the molar volume of a gas under specific conditions, we can use the Ideal Gas Law, which is expressed as:

$PV = nRT$

Where:

• P = pressure (in atm)
• V = volume (in L)
• n = number of moles
• R = Ideal Gas Constant
• T = temperature (in K)

First, we convert the temperature from Celsius to Kelvin: $T(K) = 100 + 273.15 = 373.15 \, K$

Next, we can rearrange the Ideal Gas Law to find the molar volume (V/n): $V = \frac{nRT}{P}$

For one mole of gas (n = 1), the equation simplifies to: $V = \frac{RT}{P}$

Using the Ideal Gas Constant for R = 0.0821 L·atm/(K·mol), we can substitute the values:

• R = 0.0821 L·atm/(K·mol)
• T = 373.15 K
• P = 1.00 atm

Substituting into the equation: $V = \frac{(0.0821 \, \text{L·atm/(K·mol)}) (373.15 \, K)}{1.00 \, atm}$

Calculating this gives: $V \approx 30.6116 \, L/mol$

Thus, the closest option for the molar volume of oxygen in the given conditions is A. 30.6 L.