at 25 °c, 1.00 mole of o₂ gas was found to occupy a volume of 12.5 l at a pressure of 198 kpa. what is the…

at 25 °c, 1.00 mole of o₂ gas was found to occupy a volume of 12.5 l at a pressure of 198 kpa. what is the value of the gas constant in $\frac{lcdot kpa}{molcdot k}$? $\frac{lcdot kpa}{molcdot k}$

at 25 °c, 1.00 mole of o₂ gas was found to occupy a volume of 12.5 l at a pressure of 198 kpa. what is the value of the gas constant in $\frac{lcdot kpa}{molcdot k}$? $\frac{lcdot kpa}{molcdot k}$

Answer

Explanation:

Step1: Convert temperature to Kelvin

$T = 25 + 273.15=298.15\ K$

Step2: Apply the ideal - gas law $PV = nRT$

We know that $P = 198\ kPa$, $V = 12.5\ L$, $n = 1.00\ mol$, and $T = 298.15\ K$. Rearranging the ideal - gas law for $R$ gives $R=\frac{PV}{nT}$.

Step3: Substitute the values into the formula

$R=\frac{198\ kPa\times12.5\ L}{1.00\ mol\times298.15\ K}$ $R=\frac{2475\ L\cdot kPa}{298.15\ mol\cdot K}\approx8.29\ \frac{L\cdot kPa}{mol\cdot K}$

Answer:

$8.29$