Question

A gas is in a Sealed container. The pressure is tripled and the temperature is doubled.

a. does the number of molecules within the container increase. decrease or stay the same?

b. By what factor does the number density of the gas increases.

Short answer

a. The number of molecules remains constant as the container is sealed.

b. The number density of the gas in the container increases by 1.5 times.

Step-by-step solution

Step 1. Theory of  molecule 

As per Avogadro statement, in a constant temperature and volume, all gas molecules in a closed container remains the same. As container is closed, all number of molecules are constant in the container.

Formula of density  

The equation of initial density $\left(N{D}_1\right)$is;

$\begin{array}{rcl}\left(N{D}_1\right)& =& \frac{n_1}{V_1}\\ & =& \frac{P_1}{R{T}_1}\\ & & \end{array}$

Where $n_1$is the initial number of molecules, $V_1$is initial volume, $P_1$initial pressure, and $T_1$is initial temperature.

The equation of final density $\left(N{D}_2\right)$is;

$\begin{array}{rcl}\left(N{D}_2\right)& =& \frac{n_2}{V_2}\\ & =& \frac{P_2}{R{T}_2}\\ & & \end{array}$

Where is final number of molecule, $V_2$is final volume, $P_2$is the final pressure $T_1$is the final Temperature.

$\begin{array}{rcl}\frac{\left(N{D}_2\right)}{\left(N{D}_1\right)}& =& \frac{P_2}{R{T}_2}\times \frac{R{T}_1}{P_1}\\ & =& \frac{P_2}{P_1}\times \frac{T_1}{T_2}\end{array}$

Now put the value 3 $P_{1}at{P}_{2}and2T_{1}to{T}_2$

$\begin{array}{rcl}\frac{\left(N{D}_2\right)}{\left(N{D}_1\right)}& =& \frac{3P_1}{2T_1}\times \frac{T_1}{P_1}\\ \left(N{D}_2\right)& =& \frac{3}{2}\left(N{D}_1\right)\\ & =& 1.5\left(N{D}_1\right)\end{array}$

So, the final number of density is $1.5$times.

The density of the gas in the container thus increases by 1.5 times.