Aparticular application calls for n2 g with a density of 1.80 g/l at 32 degrees c what must be the pressure of the n2 g in millimeters of mercury

1223.38 mmHg

Explanation:

Using ideal gas equation as:

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value =

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

1223.38 mmHg must be the pressure of the nitrogen gas.