A sample of neon gas occupies a volume of 461 mL atSTP. What is the volume of the neon gas when thepressure is reduced to 0.76 atm?

Question

please show work !!!!!!!!!!!!!!!!!!

Answer 1

So, STP (standard temperature and pressure)is 273K and 1 atm.

Since no mention is made of the temperature changing, we can assume the temperature is constant, and we can solve the problem with the info provided by using Boyle's law:

P1V1 = P2V2

P1 = initial pressure = 1 atm

V1 = initial volume = 461 mls

P2 = final pressure = 0.76 atm

V2 = final volume = ?

Solving for V2 we have...

V2 = P1V1/P2 = (1 atm)(461 mls) / 0.76 atm)

V2 = 607 mls

Answer 2

Hello, Yeilin,

Since the moles of gas do not change, we can use a simplified version of the gas laws:

P1V1/T1 = P2V2/T2

Where the two states are noted by the subscripts 1 and 2.

We can rearrange and isolate the unknown, P2 in this case:

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

Note how I've arranged the pressure and temperature variables as ratios. We can better understand, and predict the outcome of the change in conditions. The temperature does not change, so it cancels. That leaves the volume dependent on the change in pressure, by (P1/P2). We've reduced pressure by around 25%, so we should expect an increase in volume, by around that same amount.

V2 = (461ml)*(1 atm/0.76 atm)

V2 = 607 ml

That's about a 25% increase, so I'm content.

Bob

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