# Question c4606

Jan 29, 2017

$\text{24 L}$

#### Explanation:

NIST uses a temperature of ${20}^{\circ} \text{C}$ ("293.15 K", 68^@"F") and an absolute pressure of $\text{1 atm}$ ("14.696 psi", "101.325 kPa"). This standard is also called normal temperature and pressure (abbreviated as NTP).

Normal Temperature and Pressure

$\text{T = 293.15K}$
$P = 1 a t m$

so

The only difference between STP and NTP is that NTP conditions are at ${20}^{\circ} \text{C" = "293.15 K}$, rather than at ${0}^{\circ} \text{C" = "273.15 K}$.

Since only the temperature is different, use Charle's Law:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

We have:

${V}_{1} = 22.41 L$

${T}_{1} = 273.15 K$

V_2 = ? larr "This is the volume of 1 mol of a gas at NTP "#

${T}_{2} = 293.15 K .$

Plugging our information into Charle's Law:

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

$\implies \frac{22.41 L}{273.15 K} = {V}_{2} / \left(293.15 K\right)$

$\implies {V}_{2} = \left(22.41 L\right) \frac{293.15 K}{273.15 K} = 24.05 L$

$\text{Therefore, 1 mol of any gas will occupy about 24 L at NTP.}$

Hope this helped