Agas in a container with a fixed volume was originally at a pressure of 316 kpa and a temperature of 154 k
what is the new pressure in atm if the temperature is increased to 235 k?

482.2 kPa.

Explanation:

∵ P α T.

∴ P₁T₂ = P₂T₁.

P₁ = 316.0 kPa, T₁ = 154.0 K.

P₂ = ??? kPa, T₂ = 235.0 K.

∴ P₂ = (P₁T₂)/T₁ = (316.0 kPa)(235.0 K)/(154.0 K) = 482.2 kPa.

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 :