Calculate the change in internal energy (δe) for a system that is absorbing 35.8 kj of heat and is expanding from 8.00 to 24.0 l in volume at 1.00 atm. (remember that 101.3 j = 1 l·atm)

34179.2 J

Explanation:

According to the first law of thermodynamics:-

Where,  

E is the internal energy

q is the heat

w is the work done

From the question,

q = + 35.8 kJ  = 35800 J (+ sign as the heat is being absorbed)

The expression for the calculation of work done is shown below as:

Where, P is the pressure

is the change in volume

From the question,  

= 24.0 - 8.00 L = 16.0 L

P = 1.00 atm

Also, 1 atmL = 101.3 J

So,  

(work is done by the system)

So,

the answer is learned

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"