A container has a volume of #18 L# and holds #7 mol# of gas. If the container is compressed such that its new volume is #8 L#, how many moles of gas must be released to maintain a constant temperature and pressure?

Question

A container has a volume of #18   L# and holds #7   mol# of gas.  If the container is compressed such that its new volume is #8 L#, how many moles of gas must be released to maintain a constant temperature and pressure?

Olivia Askew · Accepted Answer

#3 8/9 mol#

Let n moles of gas to be released .

Given # V_1= "Initial volume of gas"=18L#

# V_2= "Final volume of gas"=8L#

# n_1= "Initial no mole of gas"=7 mol#

# n_2= "Final no.of moles of gas"=7-n#

As the temerature and pressure of two gases are same their volume should be proportional to their respective no. Of moles as we know from Avogadro's law.

So

#n_2/n_1=V_2/V_1#

#=>(7-n)/7=8/18#

#=>7-n=28/9# #=>n=7-28/9=35/9=3 8/9mol#

Sebastian Acosta · Answer

undefined Using the ideal gas law, we can solve for the number of moles of gas released. The ideal gas law equation is:

$$PV = nRT$$

Where:
- $P$ is the pressure (which remains constant)
- $V$ is the volume
- $n$ is the number of moles of gas
- $R$ is the gas constant
- $T$ is the temperature (which remains constant)

Given that the temperature and pressure remain constant, we can set up the following equation:

$$P_1V_1 = n_1RT$$

For the initial state:
$$V_1 = 18 \text{ L}$$
$$n_1 = 7 \text{ mol}$$

For the final state:
$$V_2 = 8 \text{ L}$$
$$n_2 = ?$$

Using the formula $P_1V_1 = P_2V_2$ (since pressure is constant), we can rearrange it to solve for $n_2$:

$$P_1V_1 = P_2V_2$$
$$n_1RT = n_2RT$$

Canceling out the gas constant and rearranging for $n_2$, we get:

$$n_2 = \frac{n_1V_1}{V_2}$$

Plugging in the values:
$$n_2 = \frac{7 \text{ mol} \times 18 \text{ L}}{8 \text{ L}}$$

$$n_2 = \frac{126 \text{ mol} \cdot \text{L}}{8 \text{ L}}$$

$$n_2 = 15.75 \text{ mol}$$

So, approximately 15.75 moles of gas must be released to maintain a constant temperature and pressure.