A 650 mL sample of neon gas exerts a pressure of 0.125 atm. If compressed until the pressure is 0.600 atm, what will be the final volume of the sample in mL?

Answer 1

We can solve this problem using the equation PV = nRT, also known as the ideal gas law. The identities of each variable is outlined below:

P: pressure (atm)

V: volume (L)

n: number of moles

R: the gas constant = 0.0821 L * atm / (mol *K)

T: temperature (K)

First, we need to find a value for what n, R and T are given the initial solution of 650 mL exerting a pressure of 0.125 atm. We do not need to find the value of each individual variable, but the value of them all multiplied because this is how they would function if we knew their individual values. We can do this by plugging these numbers into the equation as follows:

0.125 atm * 0.650 L = nRT

0.08125 atm*L = nRT

Now that we have the value of nRT, we can solve for the unknown volume given the pressure of 0.600 atm:

0.600 atm * V = 0.08125 atm*L

V = (0.08125 atm*L / 0.600 atm) = 0.135416667 L

V = 0.135416667 L ~ 0.14 L given significant digits

Final answer: 0.135416667 L ~ 0.14 L

Answer 2

Hi Sandra!

We use Boyle's Law P1*V1=P2*V2 to solve this problem. I'll plug in P1=0.125, V1=650, and P2=0.600 and then solve for V2. Remember that since V1 is mL, V2 will be mL as well.

0.125*650=0.600*V2

V2=0.125*650/0.600=140mL

I rounded to 140 because I need to answer in two significant figures. The decimal numbers have 3 significant figures but 650 only has 2, so my answer can only have 2.