Question

In a piston, the addition of 14.37 kJ of heat to a 100. g sample of a liquid at a constant temperature of 35.2 °C caused the liquid to vaporize (change to a gas). The vaporized gas expanded against an external pressure of 1.07 atm and a volume change of 11.49 L was observed. (Recall: 1 L• atm = 101.3 J} What was the change in the internal energy of the system, (AE in kJ)? (Enter your answer with two decimals places and no units.)

Answer 1

Answer : The change in internal energy of the system is 15.62 kJ

Explanation :

First we have to calculate the work done of the system.

$w=P\Delta V$

where,

w = work done

P = external pressure = 1.07 atm

$\Delta V$ = change in volume = 11.49 L

Now put all the given values on the above formula, we get:

$w=1.07atm\times 11.49L$

$w=12.3L.atm=12.3\times 101.3=1245.99J=1.25kJ$   (as per conversion)

Now we have to calculate the change in internal energy of the system.

As per first law of thermodynamic,

$\Delta E=q+w$

where,

$\Delta E$ = internal energy of the system

q = heat added to the system  = 14.37 kJ

w = work done of the system  = 1.25 kJ

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

$\Delta E=14.37+1.25$

$\Delta E=15.62kJ$

Hence, the change in internal energy of the system is 15.62 kJ

Answer 2

The change in internal energy of the system is 13.125 KJ.

From the question, we have the following information;

Heat added to the system = 14.37 kJ

Pressure of the system = 1.07 atm

Volume change of the gas = 11.49 L

From the first law of thermodynamics, we know that;

ΔU = q + w

ΔU = change in internal energy of the system

q = heat absorbed

w = work done

Since the gas expanded, the system does work so w is negative

But w = -(PΔV) = -( 1.07 atm × 11.49 L) = -12.29 L.atm

Also;

1 L• atm  = 101.3 J

-12.29 L.atm = -12.29 L.atm ×  101.3 J/ 1 L• atm

= -1245 J

Since heat is added to the system, q is positive, therefore;

ΔU = 14.37 KJ + (-1.245 KJ)

ΔU =13.125 KJ

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