describe the conditions necessary for 6.02 x 10 to the 23 power gas particles to occupy a volume of 22.4 dm3.
2007-01-18 14:19:02 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
You need to use the ideal gas equation: PV=nRT. P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. The figure you cite, 6.02*10^23 is roughly the number of molecules in a mole. The gas constant is 0.0821 L*atm/(K*mol). Your given volume is 22.4 dm^3 (22.4 L). So, rearranging, you have P/T=nR/V, or P/T=1*0.0821/22.4=0.00367. Thus, any combination of pressure (in atmospheres) divided by temperature (in Kelvin) which equals 0.00367 would satisfy your question. However, 22.4 L also happens to be the volume which one mole of gas occupies at standard temperature and pressure (STP), which is 1 atmosphere and 273.15 K (0 degrees Celsius). This is most likely the answer you are looking for.
2007-01-20 12:05:55 · answer #1 · answered by Wesley B 2 · 0⤊ 0⤋
I mole of gas will fulfill these conditions at "STP" standard temperature and pressure. i.e. 0 Celsius and 1 atm. pressure
2007-01-18 14:28:42 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
6.02*10=60.2
2007-01-18 14:22:59 · answer #3 · answered by Glowie 2 · 0⤊ 0⤋