Chapter 15, Problem 52GP

(a)

To determine

The work done by the gas.

Answer to Problem 52GP

The work done by the gas is $2.2\times {10}^5\text{J}$ .

Explanation of Solution

Given:

  $5.30\times {10}^5\text{J}$ of heat added to a gas. The volume is observed to increase from $1.9{\text{m}}^3$ to $4.1{\text{m}}^3$ .

Formula used:

Applying work done formula,

  $\mathrm{W}=\mathrm{P}\mathrm{\Delta }\mathrm{V}$

  $\mathrm{W}$ is work done of gas, P is pressure and V is volume of gas.

Calculation:

Calculate the work done of the gas

  $\begin{array}{c}\mathrm{W}=\mathrm{P}\mathrm{\Delta }\mathrm{V}\\ \mathrm{W}=\left(1\text{atm}\right)\left(2.2{\text{m}}^3\right)\\ =\left(1.01\times {10}^5\text{Pa}\right)\left(2.2{\text{m}}^{\text{3}}\right)\\ =2.2\times {10}^5\text{J}\end{array}$

Conclusion:

Therefore, the work done by the gas is $2.2\times {10}^5\text{J}$ .

(b)

To determine

The change in internal energy of the gas.

Answer to Problem 52GP

The change in internal energy of the gas is $3.1\times {10}^5\text{J}$ .

Explanation of Solution

Given:

  $5.30\times {10}^5\text{J}$ of heat added to a gas. The volume is observed to increase from $1.9{\text{m}}^3$ to $4.1{\text{m}}^3$ .

Formula used:

Applying the internal energy formula is used by

  $\mathrm{\Delta }\mathrm{U}=\mathrm{Q}-\mathrm{W}$

  $\mathrm{U}$is a internal energy, $\mathrm{Q}$ is heat of gas and $\mathrm{W}$ is work.

Calculation:

Substitute the value in the formula,

  $\begin{array}{c}\mathrm{\Delta }\mathrm{U}=\mathrm{Q}-\mathrm{W}\\ =\left(5.3\times {10}^5\text{J}\right)-\left(2.2\times {10}^5\text{J}\right)\\ =3.1\times {10}^5\text{J}\end{array}$

Conclusion:

Therefore, the change in internal energy of the gas is $3.1\times {10}^5\text{J}$ .

(c)

To determine

To plot: The graph of process of a PV diagram.

Explanation of Solution

Given:

  $5.30\times {10}^5\text{J}$ of heat added to a gas. The volume is observed to increase from $1.9{\text{m}}^3$ to $4.1{\text{m}}^3$ .

Calculation:

Consider pressure P on the y -axis and volume V on the x -axis.

The pressure P is constant at 1 atm and the volume changes from $1.9{\text{m}}^3\text{to}4.1{\text{m}}^3$

So, the graph of process of PV diagram is shown below.

  

Figure 1

The area covered under the PV diagram represents work done.

Conclusion:

Therefore, the graph of process of a PV diagram is shown in the figure 1.