Answer to Question #137729 in General Chemistry for Afwan

Question

A 6-L flask contains nitrogen gas at 32°C and 1.09atm pressure. What is the final pressure in the flask if an additional 10 g of N2 gas is added to the flask and the flask cooled to -52°C?

Answer 1

In order to solve this problem, the ideal gas law must be used:

$pV = nRT$ ,

where $p$ is the pressure, $V$ is the volume, $R$ is the gas constant , $n$ is the number of the moles and $T$ is the temperature.

In the final state, the temperature is -52°C, or 273.15-52 = 221.15 K. The volume of the flask doesn't change: $V$=6 L. The gas constant equals 0.08206 L·atm·K^-1·mol^-1. The number of the moles of the gas equals to the sum of the number of the moles that was already present in the flask plus the number of the moles contained in 10 g of N₂:

$n = n_{init}+ n_{added}$ .

The initial number of the moles can be calculated from the ideal gas law, applied to the initial state of the system:

$n_{init} = \frac{pV}{RT} = \frac{1.09\text{ atm}·6\text{ L}}{0.08206\text{ L atm K}^{-1}\text{mol}^{-1}·(32+273.15)\text{ K}}$

$n_{init} = 0.261$ mol.

The number of the moles of N₂ added to the system can be calculated using its molar mass:

$n_{added} = \frac{m}{M} = \frac{10 \text{ g}}{28.01 \text{ g/mol}} = 0.357$ mol.

Therefore, the total number of the moles is:

$n = 0.261 + 0.357 = 0.618$ mol.

Finally, the final pressure in the flask is calculated by using the ideal gas law:

$p = \frac{nRT}{V} = \frac{0.618\text{ mol}·0.08206\text{ L atm K}^{-1}\text{mol}^{-1}·221.15\text{ K}}{6 \text{ L}}$

$p = 1.87$ atm.

Answer: the final pressure in the flask equals 1.87 atm.