Question

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)

Answer 1

Answer:

34179.2 J

Explanation:

According to the first law of thermodynamics:-

$\Delta E = q + w$

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:

$w=-P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

From the question,  

$\Delta V$ = 24.0 - 8.00 L = 16.0 L

P = 1.00 atm

$w=-1.00\times 16.0\ atmL$

Also, 1 atmL = 101.3 J

So,  

$w=-1.00\times 16.0\times 101.3\ J=-1620.8\ J$ (work is done by the system)

So,

$\Delta E = +35800\ J-1620.8\ J = 34179.2\ J$