Question

find the volume of 6.022×x10 raised to 21 molecules of CO2 gas at 300k temp. and 2 bar pressure

Answer 1

as we know 
PV = nRT
given:- 
P = 2 bar = 2 x 10^5 N/m^2
T = 300 k
R = 8.314 J/(K. mol)
N = 6.022 x 10^21 

as we know 
n = N/ NA
NA = 6.022 x 10^23 
so n = 6.022 x 10^21 / 6.022 x 10^23 = 0.01 
then
2 x 10^5 x V = 0.01 x 8.314 x 300 
V = 24.942 / (2 x 10^5)
V = 12.471 x 10^-5 m^3 or 0.12471 ltr

regards

Answer 2

Answer:

The volume of $CO_{2}$ gas measured is $12.47 \times 10^{-5} L$.

Explanation:

Given,

The temperature, T = $300K$

The pressure, P = $2 bar$

Convert bar into pascal

• $1 bar = 10^{5} pascal$
• $2 bar = 2\times 10^{5} pascal$

The number of molecules =  $6.022\times 10^{21}$ molecules

As we know,

• 1 mole = $6.022\times 10^{23}$ molecules

Therefore,

• 1 molecule = $\frac{1}{6.022\times 10^{23}}$ mol
• $6.022\times 10^{21}$ molecules = $\frac{6.022\times 10^{21}}{6.022\times 10^{23}}$ mol = $10^{-2}$ mol

As we know,

• According to an ideal gas equation:
• $PV = nRT$

Here,

• P = Pressure
• V = Volume
• n = Number of moles
• R = Gas constant = $8.314 J/mol-K$
• T = Temperature

After putting the given values in the equation, we get:

• $2\times 10^{5} \times V = 10^{-2}\times 8.314 \times 300$
• $V = \frac{10^{-2}\times 8.314 \times 300}{2\times 10^{5} }$  = $12.47 \times 10^{-5} L$

Hence, the volume of $CO_{2}$ gas, V = $12.47 \times 10^{-5} L$.