A scuba diving cylinder contains compressed air at a pressure of 2.0×10^7 Pa in a volume of 12 dm^3. All of this air is released, at constant temperature, into a large empty flexible bag at atmospheric pressure, 1.0×10^5 Pa. No air is lost during the transfer. (a) Explain, in te…

Answer: ((a)) At constant temperature, the average speed of the air particles is unchanged. In the much larger volume of the bag, the same number of particles have far more space to move in, so each particle travels much further, on average, between successive collisions with the bag's surfac… ((b)) 2.4×10^3 dm^3 (i.e. 2400 dm^3, or 2.4 m^3) ((c)) The temperature of the gas must stay constant, and the mass of gas must stay fixed (no air escapes or is added during the transfer).. Part (a): Explaining the pressure drop in particle terms Since the temperature stays constant throughout the transfer, the average kinetic energy, and so the average speed, of the air particles does not change. What does change is the volu…

IGCSE Physics 0625: pV = Constant for an Expanding Gas (Scuba Cylinder)

Part (a): Explaining the pressure drop in particle terms

Since the temperature stays constant throughout the transfer, the average kinetic energy, and so the average speed, of the air particles does not change.

What does change is the volume available to the particles: the same number of particles that were confined to the small $12\text{ dm}^3$ cylinder are now spread through the much larger bag. Each particle now has to travel much further, on average, before it reaches the bag’s surface and collides with it, so collisions happen far less frequently per unit area. Since pressure comes from the frequency and force of these collisions, and the force per collision hasn’t changed (same speed), the much lower collision rate means a much lower pressure.

Part (b): Calculating the new volume

For a fixed mass of gas at constant temperature:

$p_1 V_1 = p_2 V_2$

Rearranging for $V_2$ :

$V_2 = \frac{p_1 V_1}{p_2}$

Substituting the values ( $p_1 = 2.0\times10^{7}\text{ Pa}$ , $V_1 = 12\text{ dm}^3$ , $p_2 = 1.0\times10^{5}\text{ Pa}$ ):

$V_2 = \frac{2.0\times10^{7} \times 12}{1.0\times10^{5}}$

$V_2 = \frac{2.4\times10^{8}}{1.0\times10^{5}}$

$V_2 = \boxed{2.4\times10^{3}}\text{ dm}^3$

So the air occupies $2400\text{ dm}^3$ (equivalently $2.4\text{ m}^3$ ) once released into the bag, about $200$ times its original volume, matching the roughly $200$ -fold drop in pressure from $2.0\times10^7\text{ Pa}$ to $1.0\times10^5\text{ Pa}$ .

(Check: $2.0\times10^7 \times 12 = 2.4\times10^8$ , and $1.0\times10^5 \times 2.4\times10^3 = 2.4\times10^8$ . The same product, as expected.)

Part (c): Conditions needed for pV = constant

The relationship $pV = \text{constant}$ only holds for a fixed mass of gas (no particles added or removed) held at constant temperature. Both conditions are satisfied here: no air escapes during the transfer, and the temperature is stated to stay constant.

Final answers

(a) Same particle speed (constant temperature), but far less frequent collisions with the bag’s surface in the much larger volume, so pressure decreases.
(b) New volume $=$ $2.4\times10^{3}\text{ dm}^3$ ( $2400\text{ dm}^3$ , or $2.4\text{ m}^3$ )
(c) Fixed mass of gas and constant temperature.