A gas has a pressure of 75 KPa and a volume of 33 Liters. What's the new
pressure if the volume is changed to 55 Liters?

45kPa

Explanation:

P1 = 75kPa = 75,000Pa

V1 = 33L

V2 = 55L

P2 = ?

To solve this question, we'll have to apply Boyle's law which states that the volume of a fixed mass of gas is inversely proportional to its pressure provided its temperature remains constant

Mathematically,

V = k / P, k = PV

P1×V1 = P2×V2 = P3×V3==Pn×Vn

P1 × V1 = P2 × V2

P2 = (P1 × V1) / V2

P2 = (75000 × 33) / 55

P2 = 45000Pa

P2 = 45kPa

The final pressure of the gas is 45kPa

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 :