An #8 L# container holds #6 # mol and #15 # mol of gasses A and B, respectively. Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from #310^oK# to #150 ^oK#. By how much does the pressure change?

Question

An #8 L# container holds #6 # mol and #15 # mol of gasses A and B, respectively.  Every three of molecules of gas B bind to one molecule of gas A and the reaction changes the temperature from #310^oK# to #150  ^oK#. By how much does the pressure change?

Leo Abeyta · Accepted Answer

There was a change in #57.573"atm"#

We need to find #DeltaP#, and by extension, #P_1# and #P_2#.

As per the Ideal Gas Law:

#PV=nRT#. Rearranging to solve for #P#:

#P=(nRT)/V#

Here, #V=8"L"#, #n_1=21"mol"#, #T_1=310"K"# and #R=0.0821"L atm K"^-1"mol"^-1#. Inputting:

#P_1=(21*0.0821*310)/8#

#P_1=66.809"atm"#

We now record the response that transpired:

#6"A"+15"B"rarr5"AB"_3+"A"#

So we know that #n_2=6"mol"#. Here, #T_2=150"K"#. Inputting these two new values into the Ideal Gas Law:

#P_2=(6*0.0821*150)/8#

#P_2=9.236"atm"#

We know that #DeltaP=P_2-P_1#. So here,

#DeltaP=9.236-66.809=-57.573"atm"#

Layla Ahmed · Answer

undefined To solve this problem, we can use the ideal gas law, which states:

$$PV = nRT$$

where:
- $P$ is the pressure
- $V$ is the volume
- $n$ is the number of moles of gas
- $R$ is the ideal gas constant (0.0821 L atm/mol K)
- $T$ is the temperature in Kelvin

First, let's calculate the initial pressure ($P_i$) of each gas using the ideal gas law:

For gas A:
$$P_i = \frac{n_A \cdot R \cdot T_i}{V}$$

For gas B:
$$P_i = \frac{n_B \cdot R \cdot T_i}{V}$$

Next, let's calculate the final pressure ($P_f$) of each gas at the new temperature (150 K):

For gas A:
$$P_f = \frac{n_A \cdot R \cdot T_f}{V}$$

For gas B:
$$P_f = \frac{n_B \cdot R \cdot T_f}{V}$$

Then, we can find the change in pressure ($\Delta P$) for each gas:

$$\Delta P = P_f - P_i$$

Finally, we can calculate the total change in pressure ($\Delta P_{total}$) by summing the changes in pressure for both gases.

Now, let's plug in the values and calculate:

Initial pressure ($P_i$) for gas A:
$$P_i = \frac{6 \text{ mol} \cdot 0.0821 \text{ L atm/mol K} \cdot 310 \text{ K}}{8 \text{ L}} = 15.15 \text{ atm}$$

Initial pressure ($P_i$) for gas B:
$$P_i = \frac{15 \text{ mol} \cdot 0.0821 \text{ L atm/mol K} \cdot 310 \text{ K}}{8 \text{ L}} = 37.88 \text{ atm}$$

Final pressure ($P_f$) for gas A:
$$P_f = \frac{6 \text{ mol} \cdot 0.0821 \text{ L atm/mol K} \cdot 150 \text{ K}}{8 \text{ L}} = 9.17 \text{ atm}$$

Final pressure ($P_f$) for gas B:
$$P_f = \frac{15 \text{ mol} \cdot 0.0821 \text{ L atm/mol K} \cdot 150 \text{ K}}{8 \text{ L}} = 22.93 \text{ atm}$$

Change in pressure ($\Delta P$) for gas A:
$$\Delta P_A = P_f - P_i = 9.17 \text{ atm} - 15.15 \text{ atm} = -5.98 \text{ atm}$$

Change in pressure ($\Delta P$) for gas B:
$$\Delta P_B = P_f - P_i = 22.93 \text{ atm} - 37.88 \text{ atm} = -14.95 \text{ atm}$$

Total change in pressure ($\Delta P_{total}$):
$$\Delta P_{total} = \Delta P_A + \Delta P_B = (-5.98 \text{ atm}) + (-14.95 \text{ atm}) = -20.93 \text{ atm}$$

Therefore, the pressure decreases by approximately $20.93$ atm.