Home
Class 12
CHEMISTRY
What will be the pressure of 14g N(2) (i...

What will be the pressure of 14g `N_(2)` (in atm) present in a 2 litre cylinder at 300K? 

Text Solution

AI Generated Solution

To find the pressure of 14 grams of nitrogen gas (N₂) in a 2-liter cylinder at 300 K, we can use the Ideal Gas Law equation, which is given by: \[ PV = nRT \] Where: - \( P \) = pressure (in atm) - \( V \) = volume (in liters) - \( n \) = number of moles of the gas ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Density of dry air ( only N_(2) and O_(2)) is 1.24g litre^(-1) at 760 m m and 300 K . Find the partial pressure of N_(2) gas in aire. ( Take R=(1)/(12) litre litre atm // mol K, mol. Wt. of N_(2) =28)

i. The initial pressure of PCl_(5) present in one litre vessel at 200 K is 2 atm. At equilibrium the pressure increases to 3 atm with temperature increasing to 250 . The percentage dissociation of PCl_(5) at equilibrium is

A container with a volume of 20.0 L holds N_(2)(g) and H_(2)O(l) at 300 K and 1.0 atm. The liquid water is then decomposed completely into H_(2)(g) and O_(2)(g) by any means, at constant temperature. If the final pressure becomes 1.86 atm, what was the mass of water (in gm) present initially. Neglect the initial volume of water. [Given : Vapour pressure of water at 300 K = 0.04 atm, R = 0.08 L-atm/K-mol]

The heat of combustion of H_(2)(g) at constant pressure and 300k is -280kJ mol^(-1) . What will be the heat of combustion at constant volume and at 300K ?

If a gas at a pressure of 10 atm at 300 K expands against a constant external pressure of 2 atm from a vol. 10 litres to 20 litres find work done ? [Isothermal process]