Find out the number of ideal gas molecules per cm3 that exist at a pressure of 10- 11mm of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at 27oC .

Question

Find out the number of ideal gas molecules per cm3 that exist at a pressure of 10- 11mm of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at 27oC . (A) 3.22× 1012 (B) 1.61× 1012 (C) 3.21× 106 (D) 3.22× 105

Tardigrade Admin · Accepted Answer

Using ideal gas equation in terms of number of molecules: PV=NkT where N = Number of molecules k= boltzman constant Here P=(10)- 11mmofHg,T=(.27+273.)K Putting these values in above equation we get: (N/V)=(P/k T) ⇒ (N/V)=(10- 11 × 10- 3 × 13 . 6 × 103 × 9 . 8/1 . 38 × 10- 23 × 300)=3.22× 105

Q. Find out the number of ideal gas molecules per $c m^{3}$ that exist at a pressure of $1 0^{- 11} mm$ of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at $2 7^{o} C$ .

$3.22 \times 1 0^{12}$

$1.61 \times 1 0^{12}$

$3.21 \times 1 0^{6}$

$3.22 \times 1 0^{5}$

Q. Find out the number of ideal gas molecules per cm3 that exist at a pressure of 10−11mm of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at 27oC .

3.22×1012

1.61×1012

3.21×106

3.22×105

Using ideal gas equation in terms of number of molecules : PV=NkT where N = Number of molecules k= boltzman constant Here P=(10)−11mmofHg,T=(27+273)K Putting these values in above equation we get: VN​=kTP​ ⇒VN​=1.38×10−23×30010−11×10−3×13.6×103×9.8​=3.22×105

Q. Find out the number of ideal gas molecules per $c m^{3}$ that exist at a pressure of $1 0^{- 11} mm$ of Hg, which is actually the lowest pressure (the best vacuum) that can be created in a laboratory at $2 7^{o} C$ .

$3.22 \times 1 0^{12}$

$1.61 \times 1 0^{12}$

$3.21 \times 1 0^{6}$

$3.22 \times 1 0^{5}$