A balloon of air occupies 10.0L(V2)at25.0°C(T2)and1.00atm(P2) . What temperature (T1) was it initially, if it occupied 9.40 L (V1) and was in a freezer with a pressure of 0.939 atm (P1)?

The initial temperature was 263 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

where,

= initial pressure of gas = 0.939 atm

= final pressure of gas = 1.00 atm

= initial volume of gas = 9.40 L

= final volume of gas = 10.0 L

= initial temperature of gas = ?

= final temperature of gas =

Now put all the given values in the above equation, we get:

Thus the initial temperature was 263 K

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 :