Question

A 1.40-L gaseous system absorbs 75 J of heat and expands its volume to 2.00 L against an external pressure of 1.02 atm. What is the change in internal energy for this process? 1 $L_{atm}$ = 101.3 J

Answer 1

Answer:

15.24 J is the change in internal energy for this process.

Explanation:

According to the first law of thermodynamics:-

$\Delta U = q + w$

Where,  

U is the internal energy

q is the heat

w is the work done

From the question,

q = + 75 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,  

$\Delta V$ = 2.00 - 1.40 L = 0.60 L

P = 1.02 atm

$w=-1.02\times0.60\ atmL$

Also, 1 atmL = 101.3 J

So,  

$w=-1.02\times0.60\times 101.3\ J=-59.76\ J$ (work is done by the system)

So,

$\Delta U = +75\ J-59.76\ J = 15.24\ J$

15.24 J is the change in internal energy for this process.