91Ó°ÊÓ

According to the First Law of Thermodynamics, energy can only be changed from one form to another and cannot be generated or destroyed.

According to First law of thermodynamics work done is given by

$\mathbf{W}\mathbf{=}\mathbf{Q}\mathbf{-}\mathbf{∆}\mathbf{U}$

Here, is the amount of work done by the gas, Q is the amount of heat, and$\mathbf{∆}\mathbf{U}$ is the change in internal energy.

The number of moles, n = 0.150 moles

The initial pressure,${\mathrm{T}}_1=300\mathrm{K}$

The first process, the gas is compressed.

${\mathrm{V}}_2=2{\mathrm{V}}_1$

Pressure,$\mathrm{p}=150\mathrm{Pa}$

The temperature,${\mathrm{T}}_2=150\mathrm{K}$

Calculate the work donein the first process gas is compressed isobarically. The pressure is constant in Isobaric process means

Define the work done as below.

$$\begin{gathered} \mathrm{W}=\mathrm{p}∆\mathrm{T} \\ =\mathrm{nR}∆\mathrm{T} \end{gathered}$$

Take $\mathrm{p}∆\mathrm{V}=\mathrm{nR}∆\mathrm{T}$and $∆\mathrm{T}={\mathrm{T}}_2-{\mathrm{T}}_1$

$\mathrm{W}=\mathrm{nR}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)$ ….. (1)

Here, the gas constant, $\mathrm{R}=8.314\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}·\mathrm{K}$}\right.$,

The initial and final temperature is ${\mathrm{T}}_1$and ${\mathrm{T}}_2$respectively.

Put all the known values into equation (1).

$$\begin{gathered} \mathrm{W}=0.150×8.314×\left(150-300\right) \\ =-187\mathrm{J} \end{gathered}$$

The heat added to the gas at constant pressure is,

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{p}}∆\mathrm{T}$

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{p}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)$ ….. (2)

Here, Q is the amount of heat, is number of moles,${\mathrm{C}}_{\mathrm{p}}$ is the molar heat capacity, and$∆\mathrm{T}$ is the change in temperature.

The molar heat capacity is,${\mathrm{C}}_{\mathrm{p}}=29.07\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}·\mathrm{K}$}\right.$

Put all the known values into equation (2) to getthe amount of heat.

$$\begin{gathered} \mathrm{Q}=0.150×29.07×\left(150-300\right) \\ =-645\mathrm{J} \end{gathered}$$

Here, the heatliberatedfrom the Nitrogen gas because the heat is negative.

Apply the first law of thermodynamics and calculate the change in internal energy as below.

$∆\mathrm{U}=\mathrm{Q}-\mathrm{W}$ ….. (3)

Put all the known value in the above equation, and you have

$$\begin{gathered} ∆\mathrm{U}=-645-\left(-187\right) \\ =-467\mathrm{J} \end{gathered}$$

Thus, the work done by the gas, the heat added to it, and its internal energy change during the initial compression is $\mathrm{W}=-187\mathrm{J},\mathrm{Q}=-654\mathrm{J}$and$∆\mathrm{U}=-467\mathrm{J}$ respectively.

Thesecond processexpandsadiabatically to its original volume so noheat is addedduring the process. Initial temperature state is${\mathrm{T}}_2$ and${\mathrm{T}}_3$ final state is and the work done is,

$\mathrm{W}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_3\right)$ ….. (4)

Take temperature ${\mathrm{T}}_2=150\mathrm{K}$, temperature ${\mathrm{T}}_3=114\mathrm{K}$and${\mathrm{C}}_{\mathrm{v}}$ is the molar heat capacity.

Put all the known values into equation (4).

$$\begin{gathered} \mathrm{W}=0.150×20.76×\left(150-114\right) \\ =112\mathrm{J} \end{gathered}$$

For Heat:

No heat transfer in the adiabatic process and the gas is insulated thus, the heat is zero. Therefore,

$\mathrm{Q}=0$

For the change in internal energy:

Apply the first law of thermodynamics and calculatethe change in internal energy.

Put all the known values into equation (3).

$$\begin{gathered} ∆\mathrm{U}=\mathrm{Q}-\mathrm{W} \\ =0-112 \\ =-112\mathrm{J} \end{gathered}$$

For the third process, volume is constantin the Isochoricprocess.

The work done:

At constant volume the work done is zero. Therefore,

$\mathrm{W}=0$

The heat at constant volume is,

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{v}}∆\mathrm{T}$

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}}\right)$ ….. (5)

As known,

The final temperature,${\mathrm{T}}_{\mathrm{f}}=300\mathrm{K}$

The initial temperature, ${\mathrm{T}}_{\mathrm{i}}=114\mathrm{K}$

Put all values into equation (5).

$$\begin{gathered} \mathrm{Q}=0.150×20.76×\left(300-114\right) \\ =579\mathrm{J} \end{gathered}$$

Define the change in internal energy by putting all the known value into equation (3).

$$\begin{gathered} ∆\mathrm{U}=\mathrm{Q}-\mathrm{W} \\ =579\mathrm{J}-0 \\ =579\mathrm{J} \end{gathered}$$

Thus, the work done, the heat added, and the internal energy change during the final heating is$\mathrm{W}=0,\mathrm{Q}=579\mathrm{J}$ and$∆\mathrm{U}=579\mathrm{J}$ respectively.