Verified answer

You can figure out the number of moles of gas present from the STP and volume information. 1 mole of any gas has volume 22.4 L at STP, so the amount of gas present is

(6.50 L) x (1 mole gas / 22.4 L) = 0.29 moles gas

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Then I think you have to use the ideal gas law for the second part.

You have to convert the Celsius to Kelvin.

I personally generally convert pressure into atmospheres rather than using torrs or kPa or whatever. It doesn't really matter, but I'm just used to using the R universal gas constant that has atm rather than other units.

Ideal gas law :

PV = nRT

P = (950. torr) x (1 atm / 760 torr) = 1.25 atm

V = ?

n = 0.29 moles

R = (0.08206 L•atm / mol•K)

T = 273 K ................... [ converting Celsius to Kelvin; (0 + 273) = 273 ]

Rearranging the ideal gas law equation to solve for V gives

V = (nRT / P)

= { [(0.29 moles) x (0.08206 L•atm / mol•K) x (273 K)] / (1.25 atm) }

= 5.20 L

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And ... I believe I made that harder that it had to be (lol).

The temperature doesn't change. STP is also 0.00°C .

Only the pressure and volume change, so you could use Boyle's Law

(P1V1) = (P2V2) and solve for V2.

P1 = 760 torr

V1 = 6.50 L

P2 = 950 torr

V2 = ?

V2 = [(P1V1) / P2]

= { [(760 torr) x (6.50 L)] / 950 torr }

= 5.20 L

:)

The mixed gas regulation states that: P1V1T2 = P2V2T1 Given: P1 = 80 4.5 kPa = 0.834 atm V1 = 215 cm^3 = 0.215 L T1 = 23.5 °C = 296.5 ok P2 = a million atm V2 = ? T2 = 0 °C = 273 ok V2 = P1V1T2 / P2T1 = [(0.834 atm)(0.215 L)(273 ok)] / [(a million atm)(297 ok)] = [40 9.0 atm L ok] / [297 atm ok] = 0.165 L = 165 cm^3