If i initially have a gas at a pressure of 12 atm, volume of 23 liters, and temperature of 200 k, and then i raise the pressure to 14 atm and increase the temperature to 300 k, what is the new volume of the gas?

Answer : The volume of gas will be 29.6 L

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 = 12 atm

= final pressure of gas = 14 atm

= initial volume of gas = 23 L

= final volume of gas = ?

= initial temperature of gas = 200K

= final temperature of gas = 300K

Now put all the given values in the above equation, we get the final pressure of gas.

Therefore, the new volume of gas will be 29.6 L

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.