Question

What percentage of a sample of nitrogen must be allowed to escape if its temperature, pressure and volume are changed from ${220^ \circ }C$, 3.0 atm and 1.65 L to ${110^ \circ }C$, 0.7 atm and 1.0 L respectively?

Answer 1

As pressure, volume and temperature are given use the ideal gas equation to find the initial and final moles and then calculate the percentage accordingly.
Formulas used:
-Here we have used the ideal gas law equation:
$PV = nRT$ (1)
Where, P = pressure
V = volume
N = number of moles
R = gas constant = 8.314 $J.mo{l^{ - 1}}.{K^{ - 1}}$
= 0.082057 $L.atm.{K^{ - 1}}.mo{l^{ - 1}}$
T = temperature.

-percentage of nitrogen escaped
= difference between initial and final moles / initial moles
$\% = \dfrac{{{n_i} - {n_f}}}{{{n_i}}} \times 100$ (2)

Complete Solution :
-To find out the amount of nitrogen that has escaped when we change the pressure, volume and temperature, first we need to find out the number of initial and final moles. For this we apply the ideal gas equation (1): $PV = nRT$
For ${n_i}$ (initial moles): T = ${220^ \circ }C$ = 493 K, P = 3 atm and V = 1.65 L
3 × 1.65 = ${n_i}$ × 0.082057 × 493
${n_i} = \dfrac{{3 \times 1.65}}{{0.082057 \times 493}}$
= 0.12236 moles
For ${n_f}$ (final moles): T = ${110^ \circ }C$= 383 K, P = 0.7 atm and V = 1 L
0.7 × 1 = ${n_f}$ × 0.082057 × 383
${n_f} = \dfrac{{0.7 \times 1}}{{0.082057 \times 383}}$
= 0.022273252 moles

-Now after calculating initial and final moles we will find out the % of nitrogen gas that escaped. The percentage of nitrogen gas escaped is calculated by taking the difference between initial and final moles and dividing it by initial moles. Mathematically it can be written as: $\% = \dfrac{{{n_i} - {n_f}}}{{{n_i}}} \times 100$ (2)
For this we use equation (2) = $\dfrac{{0.12236 - 0.022273252}}{{0.12236}} \times 100$
= $\dfrac{{0.100086748}}{{0.12236}} \times 100$
= 0.8179 × 100
= 81.79 % or 81.8 %
So, we can say that 81.8 % of nitrogen gas will escape out.

Note: In such questions the most common mistake made is the value of R (gas constant). The value of R should be taken according to the units of temperature, pressure and volume.
R = 8.314 $Pa.{m^3}.{K^{ - 1}}.mo{l^{ - 1}}$
R = 0.08314 $bar.L.{K^{ - 1}}.mo{l^{ - 1}}$
R = 8.314 $J.{K^{ - 1}}.mo{l^{ - 1}}$
R = 0.082057 $L.atm.{K^{ - 1}}.mo{l^{ - 1}}$