`4 g` hydrogen is mixed with 11.2 litre of He at (STP) in a container of volume 20 litre. If the final temperature is `300 K`, find the pressure.
Text Solution
Verified by Experts
4 gm Hydrogen =2 moles Hydrogen and 11.2l. He at S.T.P `=1//2` mole of He `P=P_(H)=P_(He)=(n_(H)+n_(He))(RT)/V=(2+1//2)(8.3xx(300K))/((20xx10^(-3))m^(3))=3.11xx10^(5)N//m^(2)`
FIITJEE|Exercise Numerical based Question|2 Videos
KINEMATICS
FIITJEE|Exercise NUMERICAL BASED QUESTIONS DECIMAL TYPE|5 Videos
Similar Questions
Explore conceptually related problems
12 gm He and 4 gm H_(2) is filled in a container of volume 20 litre maintained at temperature 300 K. the pressure of the mixture is nearly :
One litre of oxygen at a pressure of 1 atm and two litres of nitrogen at a pressure of 0.5 atm are introduced into a vessel of volume 1 litre. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is
11.2 litre of a gas at STP weighs 14 g The gas could not be
2 moles of each of hydrogen, carbon dioxide and chlorine are mixed in a close vessel of volume 3 litres and temperature 0^(@)C . Calculate the pressure exerted by the mixture. (R=8.31" J "mol^(-1)K^(-1))
2 moles each of hydrogen, carbon dioxide and chlorine are mixed in a close vessel of volume 3 litres and temperature 0^@ C . Calculate the pressure exerted by the mixture. ( R=8.31 J mol^(-1) K^(-1) )
Air in the cyclinder of a diesel engine is compressed to 1/15 of its initial volume. If initial temperature is 300K and initial pressure is 10^(5)Pa , find the final temperature and final pressure. Take gamma= 1.4 .
A weather balloon filled with hydrogen at 1 atm and 300K has volume equal to 12000 litres. On ascending it reaches a place where temperature is 250K and pressure is 0.5atm. The volume of the balloon is:
120 g of an ideal gas of molecular weight 40 is confirmed to a volume of 20 litre at 400 K , then the pressure of is:
FIITJEE-HEAT AND TEMPERATURE-NUMERICAL BASES QUESTIONS
