Question

An industrial hygienist collects a 5 liter bag sample at 725 mmHg and-4 degrees Celsius, if the bag is analyzed in a labortory at 760 mmHg and 25 degrees Celsius without correcting for the change in volume, what is the percent error?

Answer 1

ANSWER:

It is assumed that the sample collected by the hygienist was a gas.

So the answer can be found out by the use of combined gas-law which states that Volume (V) of a gas is directly proportional to its Temperature (T) and inversely proportional to the applied Pressure (P).

V ∝ T/P

Which means that the product of P and V when divided by T always gives a constant (C)

PV/T = C

Hence, to relate the changes that had occurred, the following equation can be derived.

P1V1/T1 = P2V2/T2 = P3V3/T3........ PnVn/Tn where n = ∝

Therefore, the change in volume can be calculated from the given data.

Let initial Temperature, Volume and Pressure be T1, V1 and P1 and the lab Temperature and Pressure be T2 and P2. Now, the V2 is

= (P1V1/T1)(T2/P2) = (725 mmHg x 5 L / 269 K) (298 K / 760 mmHg) = 5.284 L.

Hence, the change in volume of the collected sample is 0.284 L

Now, the %error is calculated as follows.

If the change in volume for 5 L of the sample is 0.284, then for 100 L is

= (0.284/5) x 100 = 5.68%