91Ó°ÊÓ

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-

$\mathbf{Q}\mathbf{=}\mathbf{Δ\pm «}\mathbf{+}\mathbf{W}$

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

1. Thenumber of moles of the gasis $\text{n}=3.00\text{″¾´Ç±ô}$
2. The specific heat at constant volume is${\text{C}}_V=6.00\text{ c²¹±ô}/\text{mol.K}$
3. The increased temperature of the gas is $\text{T}=50.0\text{‿é}$

For an isochoric process-

a.

$\begin{array}{l}{C}_V=\frac{Q}{n\mathrm{Δ}T}\\ Q=C_{V}n\mathrm{Δ}T\\ Q=6.00\text{ c²¹±ô/mol.K}×3.00\text{″¾´Ç±ô}×50.0\text{‿é}\\ \text{Q}=900\text{ c²¹±ô}\end{array}$

The heat transfer is $900\text{ c²¹±ô}$

For the constant volume, $\mathrm{Δ}V=0$

So, The value of work done is -

$\begin{array}{c}W=P\mathrm{Δ}V\\ =P×0\\ =0\end{array}$

The work done is zero.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\mathrm{Δ}{E}_{\mathrm{int}}+W \\ Q=\mathrm{Δ}{E}_{\mathrm{int}}+0 \\ \mathrm{Δ}{E}_{\mathrm{int}}=900\text{ c²¹±ô} \end{gathered}$$

The internal energy change in the gas-.$\mathrm{Δ}{E}_{\mathrm{int}}=900\text{ c²¹±ô}$

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

role="math" localid="1662442687345" $$\begin{gathered} \mathrm{Δ}K=n\frac{3}{2}kT \\ \mathrm{Δ}K=3.00\text{″¾´Ç±ô}×\frac{3}{2}×\left(2.0\text{ c²¹±ô/mol.K}\right)×(50.0\text{‿é} \end{gathered}$$

$\mathrm{Δ}K=450\text{ c²¹±ô}$

The change in the total translational kinetic energy:$\mathrm{Δ}K=450\text{ c²¹±ô}$.

The process at constant pressure:

According to the ideal gas law,

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

For constant pressure, thevalue of work doneis

$$\begin{gathered} \text{W}=\text{P}\mathrm{Δ}\text{V}=\text{nR}\mathrm{Δ}\text{T} \\ \text{W}=\left(3.0\text{″¾´Ç±ô}\right)×(2.0\text{ c²¹±ô}/\text{mol.K})×\left(50.0\text{‿é}\right) \end{gathered}$$

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

TheInternal energy change in the gas:

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

For the given values, the equation becomes-

$\begin{array}{c}\mathrm{Δ}{\text{E}}_{int}=n{C}_V\mathrm{Δ}T\\ =\left(3.0\text{″¾´Ç±ô}\right)(6.00\text{ c²¹±ô}/\text{mol}â‹\dots \text{K})\left(50.0\text{‿é}\right)\\ =900\text{ J}\end{array}$

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

$$\begin{gathered} \text{Q}=\mathrm{Δ}{\text{E}}_{int}+\text{W} \\ \text{Q}=900\text{cal}+300\text{cal} \\ \text{Q}=1200\text{ c²¹±ô} \end{gathered}$$

The heat transfer that took placeis $\text{Q}=1200\text{ c²¹±ô}$.

According to the ideal gas law,

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

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

$\begin{array}{c}W=P\mathrm{Δ}V\\ =nR\mathrm{Δ}T\\ \text{W}=\left(3.0\text{″¾´Ç±ô}\right)×(2.0\text{ c²¹±ô}/\text{mol.K})×\left(50.0\text{‿é}\right)\\ \text{W}=300\text{ c²¹±ô}\end{array}$

Thevalue of work done:$\text{W}=300\text{ c²¹±ô}$

TheInternal energy change in the gas:

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

For the given values, the equation becomes-

$\begin{array}{c}\mathrm{Δ}{E}_{\mathrm{int}}=n{C}_V\mathrm{Δ}T\\ =\left(3.0\text{″¾´Ç±ô}\right)(6.00\text{ c²¹±ô}/\text{mol}â‹\dots \text{K})\left(50.0\text{‿é}\right)\\ =900\text{ J}\end{array}$

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

The total translational kinetic energy change is given as

$\begin{array}{c}\mathrm{Δ}K=n\frac{3}{2}kT\\ =3.00\text{″¾´Ç±ô}×\frac{3}{2}×(2.0\text{ c²¹±ô}/\text{mol}â‹\dots \text{K})\left(50.0\text{‿é}\right)\\ =450\text{ c²¹±ô}\end{array}$

The total translational kinetic energy changeis found to be$450\text{ c²¹±ô}$.

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,

$\text{Q}=0\text{ c²¹±ô}$

The total heat transfer is zero joule.

j.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\mathrm{Δ}{E}_{\mathrm{int}}+W \\ W=Q−\mathrm{Δ}{E}_{\mathrm{int}} \\ W=0\text{cal}−900\text{ c²¹±ô} \\ \text{W}=−900\text{ c²¹±ô} \end{gathered}$$

Thevalue of work done is.

According to the first law of thermodynamics,

$$\begin{gathered} Q=\mathrm{Δ}{E}_{\mathrm{int}}+W \\ ocal-∆E_{int}-900cal \\ \mathrm{Δ}{E}_{\mathrm{int}}=900\text{ c²¹±ô} \end{gathered}$$

The internal energy change in the gas $\mathrm{Δ}{E}_{\mathrm{int}}=900\text{ c²¹±ô}$.

The total translational kinetic energy changeis

$$\begin{gathered} \mathrm{Δ}K=n\frac{3}{2}kT \\ \mathrm{Δ}K=3.00\text{″¾´Ç±ô}×\frac{{\displaystyle 3}}{{\displaystyle 2}}×(2.0\text{ c²¹±ô/mol}â‹\dots \text{K})\left(50.0\text{‿é}\right) \end{gathered}$$

$\mathrm{Δ}\text{K}=450\text{ c²¹±ô}$

The total translational kinetic energy changeis found to be $\mathrm{Δ}\text{K}=450\text{ c²¹±ô}$