Question

What is the volume of stank of nitrogen that contains 17 moles of nitrogen at 34 C under 12,000Pa?

Answer 1

Answer: The volume of nitrogen will be 3614.15 L

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

$PV=nRT$

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 12000 Pa = 12 kPa    (Conversion factor: 1 kPa = 1000 Pa)

V = Volume of gas = ?

n = number of moles of nitrogen = 17 moles

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the gas = $34^oC=[34+273]=307K$

Putting values in above equation, we get:

$12kPa\times V=17mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 307K\\\\V=3614.15L$

Hence, the volume of nitrogen will be 3614.15 L

Answer 2

Answer:

3626.76dm³

Explanation:

Given parameters:

Number of moles of Nitrogen in tank = 17moles

Temperature of the gas = 34°C

Pressure on the gas = 12000Pa

Unkown:

Volume of the tank, V =?

Converting the parameters to workable units:

We take the temperature from °C to Kelvin

K = 273 +  °C  = 273 + 34 = 307k

Taking the pressure in Pa to atm:

101325Pa = 1atm

12000Pa = 0.118atm

Solution:

To solve this problem, we employ the use of the ideal gas equation. The ideal gas law combines three gas laws which are the Boyle's law, Charles's law and the Avogadro's law.

            It is expressed as PV = nRT

The unknown is the Volume and we make it the subject of the formula

             V = $\frac{nRT}{P}$

Where R is called the gas constant and it is given as 0.082atmdm³mol⁻¹K⁻¹

Therefore  V = $\frac{17 x 0.082 x 307 }{0.118}$ = 3626.76dm³