a diver exhales a bubble with a volume of 250mL at a pressure of 2.4atm and a temp. of 15*C. How many gas particles are in this bubble?

If you had shown your work I could find your error.

PV = nRT or n = PV/RT
n = (2.4atm x 0.25L)/(0.08206L*atm x 288 K) = 0.0254 moles.
0.0254 moles x 6.022E23 = approximately 1.5E22 molecules.

To calculate the number of gas particles in the bubble, we need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of gas particles (in moles)
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)

First, let's convert the given values:
Volume (V) = 250 mL = 0.25 L (1 L = 1000 mL)
Pressure (P) = 2.4 atm
Temperature (T) = 15°C = 15 + 273.15 K (to convert from Celsius to Kelvin)

Now, we can plug the values into the ideal gas law equation:

(2.4 atm) * (0.25 L) = n * (0.0821 L·atm/(mol·K)) * (15 + 273.15 K)

Simplifying the equation:

0.6 atm L = n * 20.55 L·atm/(mol·K)

Now, we isolate n:

n = (0.6 atm L) / (20.55 L·atm/(mol·K))

n ≈ 0.029 moles

Finally, to convert moles to the number of gas particles, we can use Avogadro's number, which is approximately 6.022 × 10^23 particles per mole:

Number of gas particles = 0.029 moles * (6.022 × 10^23 particles/mole)

Number of gas particles ≈ 1.745 × 10^22 particles

Therefore, there are approximately 1.745 × 10^22 gas particles in the bubble.

Use PV = nRT and solve for n = number of moles.

Then remember 1 mole of a gas contains 6.022E23 molecules.

so is it 2.9 E26??

k thanks so much for your help!