Ideal Gas Questions?

Question

(a) An ideal gas occupies a volume of 1.0 cm^3 at 20ºC and atmospheric pressure. Determine the number of molecules of gas in the container.

(b) If the pressure of the 1.0-cm^3 volume is reduced to 1.0 X 10-11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container?

Answer 1

a) PV = nRT

P = 1 atm
V = 1 ml = 0.001 L
n = ?
T = 20ºC + 273.15 = 293.15 K
R = 0.082 L• atm / K • mole

n = PV/RT = (1 atm) (0.001L) / (0.082 L• atm / K • mole) (293.15 K)

= 4.16 *10^-5 moles

4.16 *10-5 moles * 6.023*10^23 molecules/mole =

= 2.51 *10^19 molecules

b) P1/n1 = P2/n2

P1 = 1 atm
n1 = 4.16 *10-5 moles
P2 = 1.0 X 10-11 Pa * (1atm/98066.5Pa) = 1.02 * 10^-16 atm

n2 = n1/P1 * P2 = 4.16 *10^-5 moles/1 atm * 1.02 * 10^-16 atm

n2 = 4.24*10^-21 moles

thats about 2556 molecules

Answer 2

Ideal Gas Equation
PV = nRT
P = Pressure in kilo Pascals (kPa)
V = volume in cubic metres (m^3)
R = the gas constant 8.31
T = temperature in Kelvin.

a). 1.0 cm^6 = 1.0 x 10^-6 m^3
100,000 kPa (atmospheric pressure).
20 degrees C = 293K
n = PV/RT
n = 100,000 x 1.0 x 10^-6 / 8.31 x 293
n = 4.10 x 10^-5 moles

Using the Avogadro Constant (6.022 x 10^23) the number of molecules in 1 mole, then:-
The gas contains = 4.10 x 10^-5 x 6.022 x 10^23 =
2.46902 x 10^19 molecules.

b).
1.0 x 10^-6 m^3
1.0 x 10^-14 kPa
293K (say)

n = 1.0 x 10^-6 x 1.0 x 10-14 / 293 x 8.31
n = 4.1 x 10^-24 moles
The gas contains 4.1x 10^-24 x 6.022 x 10^23 =
2.4732 molecules (say 2.50 molecules); your extremely good vacuum.!!!!!

Answer 3

(a)
you can calculate the moles of the gas from the ideal gas law-
n = PV/RT
n = 1 atm x 0.001 L / (0.08206 x 293.15)
n = 4.16x10^-5
molecules = 4.16 x 10^-5 x 6.022 x 10^23
molecules = 2.51 x 10^19

(b) 1.0 x 10^-11 Pa = 1.0x10^-11 / 1.0x10^5 atm
P = 1x10^-16 atm
n = 1x10^-16 atm x 0.001 L / (0.08206 x 293.15)
n = 4.2 x 10^-21 moles
molecules = 2.5 x 10^3

Answer 4

1.0 cm^3 = 1.0 x 10^-3 dm^3
p = 1 atm
T = 20 + 273 = 293 K
pV = nRT
n = pV / RT = 1 x 1x10^-3 / 0.0821 x 293 = 0.0000416 mole
Molecules = 0.0000416 x 6.02 x 10^23 = 2.50 x 10^19

1 atm = 101325 Pa
1.0 x 10^-11 / 101325 = 9.87 x 10^-17 atm
n = 9.87 x 10^-17 x 1x 10^-3 / 0.0821 x 293 = 4.10 x 10^-21 moles

Answer 5

(a)

n = PV / RT = (1.00 atm)(0.001 l) / (.0821 l atm/mole K)(293 K)

n= 4.16 Exp -5 moles

Molecules = (4.16 Exp -5 moles)(6.022 Exp 23 molecules/mole) = 2.50 Exp 19 molecules

(b) P = (1.00 Exp -11 Pa)(9.89 Exp -6 atm Pa) = 9.89 Exp -17 atm

n = PV/Rt = (9.89 Exp -17 atm)( 0.001 l) / 0.0821 l atm/mole K)(293 K) = 4.11 Exp -21 moles

molecules = (4.11 Exp -21 moles)(6.02 Exp 23 molecules / mole) = 2475 molecules


Answers: (a) 3 Exp 19 molecules (b) 2000 molecules (In both (a) and (b), you are limited to only one significant figure because there is only one significant figure in "20" degrees Celsius.

Answer 6

the quantity of gas n and the quantity V are fixed. as a result the ratio of the stress/temperature P/T is persevering with. as a result T(new) = T(previous) x P(new)/P(previous) Convert your temperature to Kelvin (absolute temperature) for this to artwork.

Answer 7

PV=nRT...check out this link

http://scienceworld.wolfram.com/physics/IdealGasLaw.html

http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html

If you have a specific question about how to do the problem then ask...