18+273 = 291 K
0 + 273 = 273 K
P V = n R T
103 * 23 = n R *291
n R = 103*23/291
same n R later
101 V = n R 273
V = n R (273/101cm^3)
V = (103*23/291)(273/101) cm^3
22 cm^3 = .0022 Liters = 22*10^-6m^3
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P V = n R T
101325 (22.4 liters)(1m^3/1000Liters) = 1(R)(273)
R = 8.313846154
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(103000)(23cm^3/10^6cm^3/m^3)=
n (8.31)(291)
n = .103*23/(8.31*291) = 9.8*10^-4 mols =.00098 mols
At a temperature of 18.0°C of 23.0cm^3 of a certain gas occupies has pressure of 103 kPa.
(a) What volume would the gas occupy at a temperature of 0.00°C and pressure of 101kPa?
(b) 1.00 moles of a certain gas occupies a volume of 22.4 litres at 0.00°C and 101325 Pa. Show that the gas constant R = 8.31 J K^-1 mol^-1.
(c) Determine the number of moles of gas in the 23.0cm^3 sample above.
1 answer
1 answer