Question

As a system increases in volume, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 atm. The final volume of the system is 58.0 L. What was the initial volume of the system if the internal energy of the system decreased by 102.5 J.

Short answer

All energy transfer can be described either as heat flow or as work done on or by the system, therefore total amount of heat goes in or outside plus total amount of work done must be equal to the internal energy.

The initial Volume of the system = 54.9 L.

Step-by-step solution

Conversion of units 

In this we will first convert the units Joules to atm. L

Joules to atm. L

$$\begin{gathered} 1L-atm=101.33joules \\ 1J=\frac{1}{101.33}L-atm \\ 52.5J=\frac{52.5}{101.33}L-atm \\ =0.51L-atm \end{gathered}$$

Also,

$$\begin{gathered} 102.5J=\frac{102.5}{101.33}L-atm \\ =1.01L-atm \end{gathered}$$

Simplification of some values

$\begin{array}{rcl}∆E& =& -102.5J\\ & =& -1.01L-atm\\ q& =& 52.5J\\ & =& 0.51L-atm\\ w& =& -P(V_2-V_1)\\ & =& -0.500(58.0-V_1)\\ & =& -29+0.500V_1\\ & & \end{array}$

Calculation of  V1

We know that heat flow or as work done on or by the system , therefore total amount of heat goes in or outside plus total amount of work done must be equal to the internal energy.

We know,$\Delta E=q+w$

Put values from step 1 and step 2 in this equation

$$\begin{gathered} -1.01=+0.51-29+0.500V_1 \\ -1.01-0.51+29=+0.500V_1 \\ 27.49=0.500V_1 \end{gathered}$$

$$\begin{gathered} V_1=\frac{27.49}{0.500} \\ =54.9L \end{gathered}$$