volume

Question

7.4 The volume of 2 grams of nitrogen at 27 °C is 0,002 m 3. Calculate the following: 7.4.1 The pressure of the gas if the gas constant for nitrogen is 297 J/kgk. 7.4.2 The final volume if the pressure doubles and the temperature increases to 125 °C.

Answer 1

The amount of nitrogen is 2 grams .

Molecular weight of nitrogen is 28 grams / mole .

Number of moles (n) = amount of nitrogen / molecular weight

                                      = 2 grams / 28 grams / mole

                                      = 0.0714 moles 

The volume of nitrogen (V) is  0.002 m³ .

Temperature T = 27° = 27+273 = 300 K

Gas constant R = 297 J/kg K

7.4.1)

The pressure of the gas can be evaluated  using ideal gas law is p = (nRT)/V .

p = (nRT)/V

p = ( 0.0714 * 297 *300 ) / 0.002

p = 3182142.857 pascal

The pressure of the nitrogen  gas is p = 3182142.857 pascal .

 

Answer 2

7.4.2)

The pressure of nitrogen gas is 3182142.857 pascal       [from 7.4.1 ]

The pressure is doubled ⇒ 2 * 3182142.857 pascal = 6364285.714 pascal .

The temperature is 125° = 125 + 273 = 398 K

The volume of nitrogen (V) is can be calculated using formula  V = (nRT)/P 

 V = (nRT)/P  

V = (0.0714 * 297 * 398) /6364285.714

V =  0.001326 m³

The volume of nitrogen (V) is 0.001326 m³ .