A $5 L$ $5 L$ container holds $15$ $15$ mol and $6$ $6$ 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 $360 K$ $360K$ to $210 K$ $210K$ . By how much does the pressure change?

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A $5 L$ $5 L$ container holds $15$ $15$ mol and $6$ $6$ 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 $360 K$ $360K$ to $210 K$ $210K$ . By how much does the pressure change?

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Answer 1 · 2016-04-05T12:42:53

The pressure changed by $8.03 \times 10^{6} P a$ $8.03 × 10^6 Pa$ .

Explanation:

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:
$\frac{p V}{n T} =$ $(pV)/(nT) =$ constant.
You may be familiar with the equation stated as: $p V = n R T$ $pV = nRT$ 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:
$\frac{p_{1} V_{1}}{n_{1} T_{1}} = \frac{p_{2} V_{2}}{n_{2} T_{2}}$ $(p_1V_1)/(n_1T_1) = (p_2V_2)/(n_2T_2)$

The volume does not change so $V_{1} = V_{2}$ $V_1=V_2$ and they will cancel out of the equation. We need to find the final pressure so rearrange for $p_{2}$ $p_2$ :
$p_{2} = p_{1} \frac{V_{1} n_{2} T_{2}}{V_{2} n_{1} T_{1}} = p_{1} \frac{n_{2} T_{2}}{n_{1} T_{1}}$ $p_2 = p_1(V_1n_2T_2)/(V_2n_1T_1) = p_1(n_2T_2)/(n_1T_1)$

The above equation will give us p₂ as a multiple of p₁. Now substitute the values into the equation:
$p_{2} = p_{1} \frac{13 \times 210}{21 \times 360} = 0.3611 p_{1}$ $p_2 = p_1(13 × 210)/(21 × 360) = 0.3611 p_1$

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:
$0.639 × 12.565 × 10^6 = 8.03 × 10^6 Pa$ .

Answer 2 · 2016-04-09T02:32:42

enter image source here
As explained in the above figure
Initial number of moles of gas molecules was $n_1=21$ moles
and Final number of moles of gas molecules was $n_2=13$ moles

We Know from Equation of state for ideal gas

$PV = nRT$ ,

where:

P = pressure in atm
V = volume in L
T = temperature in K
n = number of moles
R = universal gas constant.=0.082 $LatmK^-1mol^-1$

For Initial state
If pressure is $P_1$
Volume $V =5L$
Number of moles $n_1=21$
Temperature $T_1=360K$
and then $P_1=(n_1RT_1)/V$

For Final state
If pressure is $P_2$
Volume $V =5L$
Number of moles $n_2=13$
Temperature $T_2=210K$
and then $P_2=(n_2RT_2)/V$

So decrease in Pressure:

$= P_1-P_2=R/Vxx(n_1T_1-n_2T_2)$
$= 0.082/5xx(21xx360-13xx210)$ $atm$
$= 0.082/5xx4830$ $atm=79.212$ $atm$
$= 79.212$ $atm$

When we convert this to $Pa$ , we get:

$79.212$ $cancel(atm)xx101.325xx10^3$ $(Pa)/cancel(atm)$

$~~ 8026xx10^3$ $Pa$

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A $5 L$ $5 L$ container holds $15$ $15$ mol and $6$ $6$ 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 $360 K$ $360K$ to $210 K$ $210K$ . By how much does the pressure change?

2 Answers

Explanation:

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A 5 L5L5 L container holds 15 1515 mol and 6 66 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 360K360K360K to 210K210K210K. By how much does the pressure change?

2 Answers

Explanation:

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A $5 L$ $5 L$ container holds $15$ $15$ mol and $6$ $6$ 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 $360 K$ $360K$ to $210 K$ $210K$ . By how much does the pressure change?