1) An X-ray tube at STP is evacuated to a pressure of 1 x 10^-4 kPa at 15 degrees celcius. How many molecules?
1) An X-ray tube at STP is evacuated to a pressure of 1 x 10^-4 kPa at 15 degrees celcius. How many molecules o gas per cm^3 are there in the evacuated tube?
Answer:
PV=nRT; Rearranging, (P/RT) = (n/V)
STP = 100 kPa and 273 K.
Using the appropriate form of R:
n/V = (100 kPa) / (8.314 L-kPa / K-mol) (273 K)
n/V = 0.044 mol/L.
Converting to mol/cm^3:
n/V = (0.044 mol/L) (1 ml/cm^3) / (1000 ml/L)
n/V = 4.4 x 10^-5 mol/cm^3.
Number of molecules:
= (4.4 x 10^-5 mol/cm^3) (6.02 x 10^23 /mol)
= 2.65 x 10^19 molecules/cm^3 in the initial state.
Going from 100 kPa to 10^-4 kPa is a factor of 10^-6 (Boyle’s Law)
Going from 273 K to 288 K is a factor of 1.05 (Charles’ Law)
The number of molecules in the final state:
= (2.65 x 10^19 molecules/cm^3) (1.05) / (10^6)
= 2.80 x 10^13 molecules/cm^3.
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Answer:
PV=nRT; Rearranging, (P/RT) = (n/V)
STP = 100 kPa and 273 K.
Using the appropriate form of R:
n/V = (100 kPa) / (8.314 L-kPa / K-mol) (273 K)
n/V = 0.044 mol/L.
Converting to mol/cm^3:
n/V = (0.044 mol/L) (1 ml/cm^3) / (1000 ml/L)
n/V = 4.4 x 10^-5 mol/cm^3.
Number of molecules:
= (4.4 x 10^-5 mol/cm^3) (6.02 x 10^23 /mol)
= 2.65 x 10^19 molecules/cm^3 in the initial state.
Going from 100 kPa to 10^-4 kPa is a factor of 10^-6 (Boyle’s Law)
Going from 273 K to 288 K is a factor of 1.05 (Charles’ Law)
The number of molecules in the final state:
= (2.65 x 10^19 molecules/cm^3) (1.05) / (10^6)
= 2.80 x 10^13 molecules/cm^3.
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