Question

At STP (101.3 kPa, 0 degrees Celsius)one mole of Ar gas was collected and was found to occupy a volume of 22.4 L. What is the value of the gas constant in L*kPa/mol*K?

Answer 1

Answer:

$R_{u} = 8.307\,\frac{kPa\cdot L}{mol\cdot K}$

Explanation:

Let assume that Argon behaves ideally. The equation of state for an ideal state is:

$P\cdot V = n \cdot R_{u}\cdot T$

The gas constant is:

$R_{u} = \frac{P\cdot V}{n\cdot T}$

$R_{u} = \frac{(101.3\,kPa)\cdot (22.4\,L)}{(1\,mol)\cdot (273.15\,K)}$

$R_{u} = 8.307\,\frac{kPa\cdot L}{mol\cdot K}$