A sample of gas occupies a volume of 22.3 mL . As it expands, it does 131.5 J of work on its surroundings at a constant pressure of 777 Torr . What is the final volume of the gas?

1.29L

Explanation:

The equation that represents the change in energy for the expansion of a gas under constant pressure is:

W = PΔV

Where W is work in atmL, P is pressure in atm and ΔV is change in volume in liters

Work in atmL:

131.5J * (1atmL / 101.325J) = 1.298atmL

P in atm:

777 torr * (1atm / 760torr) = 1.022atm

ΔV is:

W / P = 1.298atmL / 1.022atm = 1.27L

As initial volume of the gas is 22.3L = 0.0223L

Final volume = 1.27L + 0.0223L =

a is the answer to this question

explanation:

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 :