Chapter 16, Problem 19SP

An ideal gas is in a chamber at a pressure of 2.00 MPa and has a volume of 20.0 liters when at a temperature of 298.15 K. Determine the number of moles of gas in the chamber.

To determine

The number of moles of gas in a chamber at a pressure of $2.0\text{ MPa }$ that has a volume of $20.0\text{ L}$ at $298.15\text{ K}$.

Answer to Problem 19SP

Solution:

$16.1\text{ mol}$

Explanation of Solution

Given data:

The pressure of the gas is $2.0\text{ MPa}$.

The volume of the ideal gas is $20.0\text{ L}$.

The temperature of the ideal gas is $298.15\text{ K}$.

Formula used:

From ideal gas law,

$PV=nRT$

Here, $P$ is the absolute pressure, $V$ is the volume of the gas, $n$ is the number of moles of the gas, $T$ is the temperature of the gas, and $R$ is the universal gas constant.

Explanation:

Recall the expression obtained from ideal gas law:

$PV=nRT$

Rearrange for number of moles of the gas:

$n=\frac{PV}{RT}$

Substitute $2.0\text{ MPa}$ for $P$, $20.0\text{ L}$ for $V$, $298.15\text{ K}$ for $T$, and $8.314\text{ J/mol}\cdot \text{K}$ for $R$

$\begin{array}{c}n=\frac{\left(2.0\text{ MPa}\right)\left(20.0\text{}\text{ L}\right)}{\left(8.314\text{ J/}\left(\text{mol}\cdot \text{K}\right)\right)\left(298.15\text{ K}\right)}\\ =\frac{\left(2.0\text{ MPa}\right)\left(\frac{{10}^6\text{}\text{Pa}}{1\text{ MPa}}\right)\left(20.0\text{}\text{ L}\right)\left(\frac{{10}^{-3}{\text{ m}}^3}{1\text{ L}}\right)}{\left(8.314\text{ J/}\left(\text{mol}\cdot \text{K}\right)\right)\left(298.15\text{ K}\right)}\\ =16.1\text{ mol}\end{array}$

Conclusion:

The number of the moles of the gas in the chamber is $16.1\text{ mol}$.