# 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?

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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