The temperature of$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{mol}\mathbf{}$of a gas with${\mathbf{C}}_v\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{cal}\mathbf{/}\mathbf{mol}\mathbf{.}\mathbf{K}\mathbf{}$ is to be raised$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{K}$ . If the process is at constant volume, what are (a) the energy transferred as heat Q, (b) the work W done by the gas, (c) the change${\mathbf{\Delta E}}_{\mathrm{int}}$ in internal energy of the gas, and (d) the change$\mathbf{\Delta K}$ in the total translational kinetic energy? If the process is at constant pressure, what are (e) Q, (f) W, (g) ${\mathbf{\Delta E}}_{\mathrm{int}}$, and (h) $\mathbf{\Delta K}$? If the process is adiabatic, what are (i) Q, (j) W, (k)${\mathbf{\Delta E}}_{\mathrm{int}}$ , and (l)$\mathbf{\Delta K}$?

We canuse the concept of the first law of thermodynamics. It states that during a process, the heat transferred to any system is partially utilized in doing work and partially in increasing the internal energy of the system. Mathematically, it can be represented as-

$\mathbf{Q}\mathbf{=}\mathbf{\Delta U}\mathbf{+}\mathbf{W}$

Here Q is the heat transferred, $\mathbf{\Delta U}$ is the change in internal energy and W is the work done.

1. Thenumber of moles of the gasis $\text{n}=3.00\text{\hspace{0.17em}mol}$
2. The specific heat at constant volume is${\text{C}}_V=6.00\text{\hspace{0.17em}cal}/\text{mol.K}$
3. The increased temperature of the gas is $\text{T}=50.0\text{\hspace{0.17em}K}$

For an isochoric process-

a.

$\begin{array}{l}{C}_V=\frac{Q}{n\Delta T}\\ Q=C_{V}n\Delta T\\ Q=6.00\text{\hspace{0.17em}cal/mol.K}\times 3.00\text{\hspace{0.17em}mol}\times 50.0\text{\hspace{0.17em}K}\\ \text{Q}=900\text{\hspace{0.17em}cal}\end{array}$

The heat transfer is $900\text{\hspace{0.17em}cal}$

For the constant volume, $\Delta V=0$

So, The value of work done is -

$\begin{array}{c}W=P\Delta V\\ =P\times 0\\ =0\end{array}$

The work done is zero.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\Delta E_{\mathrm{int}}+W \\ Q=\Delta E_{\mathrm{int}}+0 \\ \Delta E_{\mathrm{int}}=900\text{\hspace{0.17em}cal} \end{gathered}$$

The internal energy change in the gas-.$\Delta E_{\mathrm{int}}=900\text{\hspace{0.17em}cal}$

The expression of the change in the total translational kinetic energy is

role="math" localid="1662442687345" $$\begin{gathered} \Delta K=n\frac{3}{2}kT \\ \Delta K=3.00\text{\hspace{0.17em}mol}\times \frac{3}{2}\times \left(2.0\text{\hspace{0.17em}cal/mol.K}\right)\times (50.0\text{\hspace{0.17em}K} \end{gathered}$$

$\Delta K=450\text{\hspace{0.17em}cal}$

The change in the total translational kinetic energy:$\Delta K=450\text{\hspace{0.17em}cal}$.

The process at constant pressure:

According to the ideal gas law,

$\text{PV}=\text{nRT}$

For constant pressure, thevalue of work doneis

$$\begin{gathered} \text{W}=\text{P}\Delta \text{V}=\text{nR}\Delta \text{T} \\ \text{W}=\left(3.0\text{\hspace{0.17em}mol}\right)\times (2.0\text{\hspace{0.17em}cal}/\text{mol.K})\times \left(50.0\text{\hspace{0.17em}K}\right) \end{gathered}$$

$\text{W}=300\text{cal}$

TheInternal energy change in the gas:

$\Delta {\text{E}}_{int}=n{C}_V\Delta T$

For the given values, the equation becomes-

$\begin{array}{c}\Delta {\text{E}}_{int}=n{C}_V\Delta T\\ =\left(3.0\text{\hspace{0.17em}mol}\right)(6.00\text{\hspace{0.17em}cal}/\text{mol}\cdot \text{K})\left(50.0\text{\hspace{0.17em}K}\right)\\ =900\text{\hspace{0.17em}J}\end{array}$

Using the first law of thermodynamics and the calculated values of work and internal energy, we get

$$\begin{gathered} \text{Q}=\Delta {\text{E}}_{int}+\text{W} \\ \text{Q}=900\text{cal}+300\text{cal} \\ \text{Q}=1200\text{\hspace{0.17em}cal} \end{gathered}$$

The heat transfer that took placeis $\text{Q}=1200\text{\hspace{0.17em}cal}$.

According to the ideal gas law,

$\text{PV}=\text{nRT}$

If pressure is constant, the value of work done by the gas is-

$\begin{array}{c}W=P\Delta V\\ =nR\Delta T\\ \text{W}=\left(3.0\text{\hspace{0.17em}mol}\right)\times (2.0\text{\hspace{0.17em}cal}/\text{mol.K})\times \left(50.0\text{\hspace{0.17em}K}\right)\\ \text{W}=300\text{\hspace{0.17em}cal}\end{array}$

Thevalue of work done:$\text{W}=300\text{\hspace{0.17em}cal}$

TheInternal energy change in the gas:

$\Delta {\text{E}}_{int}=n{C}_V\Delta T$

For the given values, the equation becomes-

$\begin{array}{c}\Delta E_{\mathrm{int}}=n{C}_V\Delta T\\ =\left(3.0\text{\hspace{0.17em}mol}\right)(6.00\text{\hspace{0.17em}cal}/\text{mol}\cdot \text{K})\left(50.0\text{\hspace{0.17em}K}\right)\\ =900\text{\hspace{0.17em}J}\end{array}$

Internal energy change in the gas is $900\text{\hspace{0.17em}J}$.

The total translational kinetic energy change is given as

$\begin{array}{c}\Delta K=n\frac{3}{2}kT\\ =3.00\text{\hspace{0.17em}mol}\times \frac{3}{2}\times (2.0\text{\hspace{0.17em}cal}/\text{mol}\cdot \text{K})\left(50.0\text{\hspace{0.17em}K}\right)\\ =450\text{\hspace{0.17em}cal}\end{array}$

The total translational kinetic energy changeis found to be$450\text{\hspace{0.17em}cal}$.

The process is adiabatic:

The process is adiabatic, hence, there is no change of energy through the system to the surrounding or vice versa. Therefore,

$\text{Q}=0\text{\hspace{0.17em}cal}$

The total heat transfer is zero joule.

j.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\Delta E_{\mathrm{int}}+W \\ W=Q-\Delta E_{\mathrm{int}} \\ W=0\text{cal}-900\text{\hspace{0.17em}cal} \\ \text{W}=-900\text{\hspace{0.17em}cal} \end{gathered}$$

Thevalue of work done is.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\Delta E_{\mathrm{int}}+W \\ ocal-∆E_{int}-900cal \\ \Delta E_{\mathrm{int}}=900\text{\hspace{0.17em}cal} \end{gathered}$$

The internal energy change in the gas $\Delta E_{\mathrm{int}}=900\text{\hspace{0.17em}cal}$.

The total translational kinetic energy changeis

$$\begin{gathered} \Delta K=n\frac{3}{2}kT \\ \Delta K=3.00\text{\hspace{0.17em}mol}\times \frac{{\displaystyle 3}}{{\displaystyle 2}}\times (2.0\text{\hspace{0.17em}cal/mol}\cdot \text{K})\left(50.0\text{\hspace{0.17em}K}\right) \end{gathered}$$

$\Delta \text{K}=450\text{\hspace{0.17em}cal}$

The total translational kinetic energy changeis found to be $\Delta \text{K}=450\text{\hspace{0.17em}cal}$