what is the volume of 0.200 mol of an ideal gas at 200. kpa and 400. k? use $pv = nrt$ and $r =…

what is the volume of 0.200 mol of an ideal gas at 200. kpa and 400. k? use $pv = nrt$ and $r = 8.314\frac{lcdot kpa}{molcdot k}$. 0.83 l 3.33 l 5.60 l 20.8 l
Answer
Explanation:
Step1: Rearrange the ideal - gas law for volume.
Given $PV = nRT$, we can solve for $V$ as $V=\frac{nRT}{P}$.
Step2: Substitute the given values.
We have $n = 0.200\ mol$, $R=8.314\frac{L\cdot kPa}{mol\cdot K}$, $T = 400\ K$, and $P = 200\ kPa$. $V=\frac{0.200\ mol\times8.314\frac{L\cdot kPa}{mol\cdot K}\times400\ K}{200\ kPa}$
Step3: Calculate the volume.
First, calculate the numerator: $0.200\times8.314\times400 = 665.12$. Then, divide by the denominator: $V=\frac{665.12}{200}=3.3256\approx3.33\ L$.
Answer:
3.33 L