A container has a volume of #4 L# and holds #3 mol# of gas. If the container is expanded such that its new volume is #12 L#, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

Question

A container has a volume of #4   L# and holds #3   mol# of gas.  If the container is expanded such that its new volume is #12 L#, how many moles of gas must be injected into the container to maintain a constant temperature and pressure?

Layla Ahmed · Accepted Answer

If a container has a constant temperature and pressure this means it has a constant concentration.

In formulaic words: #c_i = c_f# Hence, since: #c = n/v# #n_i/v_i = n_f/v_f# #3/4 = n_f/12# #n_f = 9# #mol# Therefore, #6# #mol# of gas must be added.

Nora Arzate · Answer

undefined The ideal gas law, $PV = nRT$, can be used to solve this problem. Since the temperature and pressure are constant, we can use the equation $P_1V_1 = P_2V_2$ to find the initial pressure. Then, using the ideal gas law, we can find the initial number of moles. Finally, we use the same equation with the new volume to find the final number of moles. Subtracting the initial moles from the final moles gives us the number of moles to be injected.

$P_1V_1 = P_2V_2$

$P_1 = \frac{{P_2V_2}}{{V_1}}$

$P_1 = \frac{{1 \times 12}}{{4}} = 3 \, atm$

$n_1 = \frac{{P_1V_1}}{{RT}} = \frac{{3 \times 4}}{{0.0821 \times 298}} \approx 0.49 \, mol$

$n_2 = \frac{{P_2V_2}}{{RT}} = \frac{{1 \times 12}}{{0.0821 \times 298}} \approx 0.49 \, mol$

$n_{\text{injected}} = n_2 - n_1 = 0.49 - 0.49 = 0 \, mol$

Therefore, 0 moles of gas must be injected into the container to maintain constant temperature and pressure.