Question

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6 to 10.00 L . During this process, it absorbs 500Jof heat from the surroundings.

(a) Calculate the energy change of the gas,$\mathit{\Delta }\mathit{U}$ .

(b) Calculate the work,$\mathit{w}$ , done on the gas in an irreversible adiabatic (q=0) process connecting the same initial and final states.

Short answer

a) The energy change of the gas is $-311J$.

b) Work done in the irreversible adiabatic process is $-311\text{J}$.

Step-by-step solution

Given data

The constant external pressure gas,$P_{\text{ext}}=2.00\text{atm}$.

The Initial volume,$V_2=6.00L$.

The final volume, $V_2=6.00L$.

Concept of the First law of thermodynamics

According to the First Law of Thermodynamics the change in internal energy of the system is equal to the sum of work done and heat absorbed by the system. The first Law of Thermodynamics is also known as the Law of conservation of energy. The mathematical expression for the First Law of Thermodynamics is given below.

$\mathbf{∆}\mathit{U}\mathbf{}\mathbf{=}\mathbf{}\mathit{q}\mathbf{}\mathbf{+}\mathbf{}\mathit{w}$ ……(1)

Where,

$\mathbf{∆}\mathit{U}$ change in internal energy

W = work done

Q= heat

Calculation of work done

Work done will be calculated by the formula,

$\text{w}=-{\text{P}}_{\text{ex}}\Delta \text{V}$. …… (2)

Where,$P_{ext}$ is external pressure,$\Delta V$ is the change in volume.

Now, substitute the value of$P_{ext}$and$\Delta V$ in equation (2).

$$\begin{gathered} w=-(2.00\text{atm})\times (10.00\text{L}-6.00\text{L})\times \frac{101.325\text{J}}{1\text{Latm}} \\ =-811\text{J} \end{gathered}$$

(a) Calculation of energy change

Energy change can be determined with the help of the first law of thermodynamics relation

Substitute the value of$q=500 \text{J}$and$w=-811 \text{J}$in equation (1).

$$\begin{gathered} \Delta U=+500\text{J}+(-811\text{J}) \\ =-311\text{J} \end{gathered}$$

Therefore, the energy change of the gas, $\Delta U=311\text{J}$.

(b) Calculation of work done in adiabatic condition

It is known that, for an adiabatic process$q=0$.

$$\begin{gathered} \Delta U=q+w \\ -311\text{J}=0+w \\ w=-311\text{J} \end{gathered}$$

Therefore, work done in irreversible adiabatic process is $-311\text{J}$.