A sample of a gas at room temperature occupies a volume of 40.0 L at a pressure of 262 torr. If the pressure changes to 1310 torr, with no change in the temperature or moles of gas, what is the new volume, V2

a) p1v1= p2v2

852 x 40.0= 4260 x v2

v2= 8 L

b) 852 x 40 = p2 x 69.0

p2= 494 torr

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the answer is minerals.