The pressure of a sample of argon gas was increased from 3.71 atm to 8.55 atm at constant temperature. If the final volume of the argon sample was 16.1 L, what was the initial volume of the argon sample? Assume ideal behavior.

I got 37

Use p1v1 = p2v2

You're allowed 3 places so the answer is 37.x

To solve this problem, we can use the combined gas law, which relates the initial and final pressure, volume, and temperature of a gas sample.

The combined gas law is given by the equation:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

In this case, since the temperature is constant, we can eliminate it from the equation, and the equation becomes:

(P1 * V1) = (P2 * V2)

Now we can substitute the given values into the equation:

(3.71 atm * V1) = (8.55 atm * 16.1 L)

To find the initial volume (V1), we need to rearrange the equation as follows:

V1 = (P2 * V2) / P1
= (8.55 atm * 16.1 L) / (3.71 atm)

Calculating this expression gives us:

V1 = 37.102 L

Therefore, the initial volume of the argon sample was approximately 37.102 L.