A #5 L# container holds #15 # mol and #6 # mol of gasses A and B, respectively. Every three of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from #360K# to #210K#. By how much does the pressure change?

Question

A #5 L# container holds #15 # mol and #6 # mol of gasses A and B, respectively.  Every three of molecules of gas B bind to two molecule of gas A and the reaction changes the temperature from #360K# to #210K#. By how much does the pressure change?

Olivia Askew · Accepted Answer

<img src="https://tutor.hix.ai 
As explained in the above figure 
Initial number of moles of gas molecules was <mathjax>#n_1=21 #</mathjax>moles 
and Final number of moles of gas molecules was <mathjax>#n_2=13#</mathjax> moles
We Know from Equation of state for ideal gas
<blockquote>
<mathjax>#PV = nRT#</mathjax>,
where:
<ul>
<li>P = pressure in atm</li>
<li>V = volume in L</li>
<li>T = temperature in K</li>
<li>n = number of moles </li>
<li>R = universal gas constant.=0.082<mathjax>#LatmK^-1mol^-1#
</ul>
For Initial state 
If pressure is <mathjax>#P_1#</mathjax> 
Volume <mathjax>#V =5L#</mathjax> 
Number of moles <mathjax>#n_1=21 #</mathjax> 
Temperature <mathjax>#T_1=360K#</mathjax> 
and then <mathjax>#P_1=(n_1RT_1)/V#
For Final state 
If pressure is <mathjax>#P_2#</mathjax> 
Volume <mathjax>#V =5L#</mathjax> 
Number of moles <mathjax>#n_2=13 #</mathjax> 
Temperature <mathjax>#T_2=210K#</mathjax> 
 and then <mathjax>#P_2=(n_2RT_2)/V#
So decrease in Pressure:
<mathjax>#= P_1-P_2=R/Vxx(n_1T_1-n_2T_2)#</mathjax> 
<mathjax>#= 0.082/5xx(21xx360-13xx210)#</mathjax> <mathjax>#atm#</mathjax> 
<mathjax>#= 0.082/5xx4830#</mathjax> <mathjax>#atm=79.212#</mathjax> <mathjax>#atm#</mathjax> 
<mathjax>#= 79.212#</mathjax> <mathjax>#atm#
When we convert this to <mathjax>#Pa#</mathjax>, we get:
<mathjax>#79.212#</mathjax> <mathjax>#cancel(atm)xx101.325xx10^3#</mathjax> <mathjax>#(Pa)/cancel(atm)#
<mathjax>#~~ 8026xx10^3#</mathjax> <mathjax>#Pa#

Benjamin Asiedu · Answer

The pressure changed by #8.03 × 10^6 Pa#.

To get started we’ll work out how many moles of gas are present at the end of the reaction. There are 6 moles of gas B. In the reaction gas B bonds with gas A at a ratio of 3:2 molecules, so that ratio will apply to the number of moles also. 3 moles of gas B will bond with 2 moles of gas A. In total all 6 moles of gas B will bond with 4 moles of gas A and the product of that reaction will be 2 moles of a new gas. Let’s call the new gas, gas C. That leaves us with 15 – 4 = 11 moles of gas A, 0 moles of gas B (it all reacted), and 2 moles of gas C. So the final number of moles of gas will be 13 moles. The initial number for moles was 15 + 6 = 21 moles. Calculate the Final Pressure To solve the problem of the pressure change we will use the ideal gas equation: #(pV)/(nT) =# constant. You may be familiar with the equation stated as: #pV = nRT# where R is the molar gas constant. Since the LHS of the equation is equal to a constant we can now state the equation like this: #(p_1V_1)/(n_1T_1) = (p_2V_2)/(n_2T_2)# The volume does not change so #V_1=V_2# and they will cancel out of the equation. We need to find the final pressure so rearrange for #p_2#: #p_2 = p_1(V_1n_2T_2)/(V_2n_1T_1) = p_1(n_2T_2)/(n_1T_1)# The above equation will give us p₂ as a multiple of p₁. Now substitute the values into the equation: #p_2 = p_1(13 × 210)/(21 × 360) = 0.3611 p_1# Pressure Change To calculate by how much the pressure changes find the difference between initial and final pressures: #p_1 – p_2 = p_1 – 0.3611 p_1 = 0.639 p_1# Calculate the value of p₁ by using the ideal gas equation: Volume conversion: # V = 5 L = 5 × 10^(-3) m³# #p_1V_1 = n_1RT_1 ⇒ p_1 = (n_1RT_1) / V_1 = (21 × 8.31 × 360) / (5 × 10^(-3)) = 12.565 × 10^6 Pa# The pressure therefore changed by: #0.639 × 12.565 × 10^6 = 8.03 × 10^6 Pa#.

Elena Albarran · Answer

undefined To solve this problem, we first need to calculate the initial pressure of the gases using the ideal gas law:

$PV = nRT$

Where:
- $P$ = pressure
- $V$ = volume (5 L)
- $n$ = number of moles of gas (15 mol for gas A, 6 mol for gas B)
- $R$ = ideal gas constant (0.0821 L·atm/(mol·K))
- $T$ = temperature in Kelvin (360 K)

Using the ideal gas law, we can find the initial pressure $P_{\text{initial}}$ of the gases.

Then, we need to find the final pressure $P_{\text{final}}$ of the gases at the new temperature (210 K). Since the number of moles and volume remain constant, we can use the ideal gas law again with the new temperature.

Once we have both initial and final pressures, we can calculate the pressure change using the formula:

$\Delta P = P_{\text{final}} - P_{\text{initial}}$

Now, let's calculate:

Initial pressure ($P_{\text{initial}}$):
$P_{\text{initial}} = \frac{n_{\text{total}}RT}{V}$

Final pressure ($P_{\text{final}}$):
$P_{\text{final}} = \frac{n_{\text{total}}RT_{\text{new}}}{V}$

Pressure change ($\Delta P$):
$\Delta P = P_{\text{final}} - P_{\text{initial}}$

Substitute the values and solve for $\Delta P$.

Elena Albarran · Answer

undefined To determine how much the pressure changes due to the given reaction, we first need to consider 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, and
- $ T $ is the temperature.

Since the volume and the number of moles of gas do not change, we can rewrite the ideal gas law as:

$$ P_1 = \frac{nRT_1}{V} $$
$$ P_2 = \frac{nRT_2}{V} $$

Where:
- $ P_1 $ is the initial pressure (at 360K),
- $ P_2 $ is the final pressure (at 210K),
- $ T_1 $ is the initial temperature (360K), and
- $ T_2 $ is the final temperature (210K).

We need to calculate $ P_1 $ and $ P_2 $ to find the pressure change. Given that $ V $, $ n $, and $ R $ are constants, we can solve for $ P_1 $ and $ P_2 $ using the ideal gas law.

After finding $ P_1 $ and $ P_2 $, the pressure change ($ \Delta P $) can be calculated as:

$$ \Delta P = P_2 - P_1 $$

Substituting the values and solving will give us the pressure change resulting from the temperature change caused by the reaction between gases A and B.