At a temperature of -33.0 degrees C, a sample of confined gas exerts a pressure of 53.3 kPa. If volume remains constant, at what temperature will the pressure reach 133 kPa?

598.9K = 325.9°C

Explanation:

P1= 53.3kPa T1= -33+273=240K

P2=133kPa , T2= ?

Applying P1/T1 = P2/T2

Substitute and Simplify

53.3/240 = 133/T2

T2= 598.9K = 325.9°C

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.