An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10 –3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 10 23 mol –1 , density of mercury = 13600 kg m –3 and g = 10 m s –2 .

Question

Philoid · Accepted Answer

We know ideal gas equation

PV=nRT

Where V= volume of gas

R=gas constant

T=temperature

n=number of moles of gas

P=pressure of gas.

Given

Volume of gas=250cc

1cc=1cm³m³

V=250m³

Pressure P=10^–3 mm of mercury

1mm of Hg= 133.32Pa

P= mm of Hg=133.32 Pa

Temperature T=27

T(K)=T ()+273.15

T=T(K)=27+273.15=300.15K

From ideal gas equation, we can write

Number of molecules = Avogadro numbernumber of moles

Number of molecules=n

=

Number of molecules in electric bulb=.

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10–3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 1023 mol–1, density of mercury = 13600 kg m–3 and g = 10 m s–2.

An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10^–3 mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 × 10²³ mol^–1, density of mercury = 13600 kg m^–3 and g = 10 m s^–2.