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?

Apr 5, 2016

The pressure changed by 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:
$\frac{p V}{n T} =$ constant.
You may be familiar with the equation stated as: $p V = n R T$ 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:
$\frac{{p}_{1} {V}_{1}}{{n}_{1} {T}_{1}} = \frac{{p}_{2} {V}_{2}}{{n}_{2} {T}_{2}}$

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 ${p}_{2}$:
${p}_{2} = {p}_{1} \frac{{V}_{1} {n}_{2} {T}_{2}}{{V}_{2} {n}_{1} {T}_{1}} = {p}_{1} \frac{{n}_{2} {T}_{2}}{{n}_{1} {T}_{1}}$

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.

Apr 9, 2016 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

$P V = n R T$,

where:

• P = pressure in atm
• V = volume in L
• T = temperature in K
• n = number of moles
• R = universal gas constant.=0.082$L a t m {K}^{-} 1 m o {l}^{-} 1$

For Initial state
If pressure is ${P}_{1}$
Volume $V = 5 L$
Number of moles ${n}_{1} = 21$
Temperature ${T}_{1} = 360 K$
and then ${P}_{1} = \frac{{n}_{1} R {T}_{1}}{V}$

For Final state
If pressure is ${P}_{2}$
Volume $V = 5 L$
Number of moles ${n}_{2} = 13$
Temperature ${T}_{2} = 210 K$
and then ${P}_{2} = \frac{{n}_{2} R {T}_{2}}{V}$

So decrease in Pressure:

$= {P}_{1} - {P}_{2} = \frac{R}{V} \times \left({n}_{1} {T}_{1} - {n}_{2} {T}_{2}\right)$
$= \frac{0.082}{5} \times \left(21 \times 360 - 13 \times 210\right)$ $a t m$
$= \frac{0.082}{5} \times 4830$ $a t m = 79.212$ $a t m$
$= 79.212$ $a t m$

When we convert this to $P a$, we get:

$79.212$ $\cancel{a t m} \times 101.325 \times {10}^{3}$ $\frac{P a}{\cancel{a t m}}$

$\approx 8026 \times {10}^{3}$ $P a$