Question

1. the volume of one mole of an ideal gas, at 2.0 atmospheres and 273 k, is a) 5.60 l b) 11.2 l c) 22.4 l d) 44.8 l e) 56.0 l

Explanation:

Step1: Recall ideal - gas law

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant ($R=0.0821\ L\cdot atm/(mol\cdot K)$), and $T$ is temperature.
We are given that $n = 1\ mol$, $P = 2.0\ atm$, $T=273\ K$, and we need to find $V$.

Step2: Rearrange the ideal - gas law for volume

From $PV=nRT$, we can solve for $V$: $V=\frac{nRT}{P}$.

Step3: Substitute the given values

Substitute $n = 1\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$, $T = 273\ K$, and $P = 2.0\ atm$ into the formula:
$V=\frac{1\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times273\ K}{2.0\ atm}$.
First, calculate the numerator: $1\times0.0821\times273 = 22.4133\ L\cdot atm$.
Then, divide by the denominator: $V=\frac{22.4133\ L\cdot atm}{2.0\ atm}=11.20665\ L\approx11.2\ L$.

Answer:

B. 11.2 L