what is the pressure of 0.540 mol of an ideal gas at 35.5 l and 223 k? use $pv = nrt$ and $r =…

what is the pressure of 0.540 mol of an ideal gas at 35.5 l and 223 k? use $pv = nrt$ and $r = 8.314\frac{lcdot kpa}{molcdot k}$. 0.715 kpa 2.45 kpa 28.2 kpa 62.7 kpa

what is the pressure of 0.540 mol of an ideal gas at 35.5 l and 223 k? use $pv = nrt$ and $r = 8.314\frac{lcdot kpa}{molcdot k}$. 0.715 kpa 2.45 kpa 28.2 kpa 62.7 kpa

Answer

Explanation:

Step1: Rearrange the ideal - gas law for pressure

We start with $PV = nRT$. Solving for $P$ gives $P=\frac{nRT}{V}$.

Step2: Substitute the given values

We are given $n = 0.540\ mol$, $R=8.314\frac{L\cdot kPa}{mol\cdot K}$, $T = 223\ K$ and $V = 35.5\ L$. Substitute these values into the formula: $P=\frac{0.540\ mol\times8.314\frac{L\cdot kPa}{mol\cdot K}\times223\ K}{35.5\ L}$. First, calculate the numerator: $0.540\times8.314\times223 = 0.540\times1854.022=991.17188$. Then, divide by the denominator: $P=\frac{991.17188}{35.5}\ kPa\approx28.2\ kPa$.

Answer:

28.2 kPa