Question

a glass tube of volume 112 ml containing a gas is partially evacuated till the pressure in it drops to 3.8*10 power-5 torr at 0 degree Celsius the number of molecules of the gas remaining in the tube is

Answer 1

Answer: The number of gas molecules remaining in the tube are $1.5055\times 10^{14}$

Explanation:

The equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the gas = $3.8\times 10^{-5}Torr$

V = Volume of the gas = 112 mL = 0.112 L     (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas = $0^oC=[0+273]K=273K$

R = Gas constant = $62.364\text{ L. torr }mol^{-1}K^{-1}$

n = number of moles of gas = ?

Putting values in above equation, we get:

$3.8\times 10^{-5}torr\times 0.112L=n_{mix}\times 62.364\text{ L torr }mol^{-1}K^{-1}\times 273K\\n_{gas}=\frac{3.8\times 10^{-5}\times 0.112}{62.364\times 273}=2.50\times 10^{-10}mol$

According to mole concept:

1 mole of a compound contains $6.022\time 10^{23}$ number of molecules

So, $2.50\times 10^{-10}mol$ of gas will contain $\frac{6.022\times 10^{23}}{1}\times 2.50\times 10^{-10}=1.5055\times 10^{14}$ number of molecules.

Hence, the number of gas molecules remaining in the tube are $1.5055\times 10^{14}$

Answer 2

Answer:1.5×10*17

Explanation:

Pv=nrt

n=pv/rt

P=. 1atm=760torque

Xatm=3.8×10*-5

then x=5×10*-5

n=5×10*-5×0.112liters÷0.0821×273k

=0.025×10*-5

0.025×10*-5×6.022×10*23

0.15×10*18

=1.5×10*17