Thermodynamics

For an adiabatic system, Q = 0

By the first law of thermodynamics

$$\begin{gathered} ∆\mathrm{Q} = ∆\mathrm{W} + ∆\mathrm{U} \\ \mathrm{Here}, ∆\mathrm{Q} = 0 \\ \mathrm{So}, ∆\mathrm{W} = - ∆\mathrm{U} \end{gathered}$$

Here, chnage in internal energy is teh function of temperature, so work done by the gas depends on the chnage in temperature.

The equilibrium state of a thermodynamic system is described by macroscopic state variables like Pressure (P), Temperature (T), and Volume (V). Heat and Work are energy transfers, not state variables.

(NEW NCERT 11^th Page No. 227, 228)

The equilibrium state of a thermodynamic system is described by macroscopic state variables like Pressure (P), Temperature (T), and Volume (V). Heat and Work are energy transfers, not state variables.

(NEW NCERT 11^th Page No. 227, 228)

From the first law of thermodynamics for an isobaric process: $$\Delta Q = \Delta U + \Delta W.$$

• For a monoatomic gas:
  • $\Delta U = \frac{3}{2} n R \Delta T$,
  • $\Delta W = n R \Delta T$.
• Heat supplied: $$\Delta Q = \frac{3}{2} n R \Delta T + n R \Delta T = \frac{5}{2} n R \Delta T.$$
• Ratio: $$\frac{\Delta Q}{\Delta U} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3}.$$

Thus, the ratio is $\frac{5}{3}$.

The adiabatic relation between the temperature and volume for an ideal gas is:

By the equation of Adiabatic Process:

$$\begin{gathered} {\mathrm{T}}_{\mathrm{f}}{\mathrm{V}}_{\mathrm{f}}^{\mathrm{\gamma }-1} = {\mathrm{T}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}^{\mathrm{\gamma }-1} \\ \mathrm{Subtitute}, {\mathrm{V}}_{\mathrm{f}} = \frac{1}{4}{\mathrm{V}}_{\mathrm{i}} \\ {\mathrm{T}}_{\mathrm{f}}{\left(\frac{{\mathrm{V}}_{\mathrm{i}}}{4}\right)}^{\mathrm{\gamma }-1} = {\mathrm{T}}_{\mathrm{i}}{\mathrm{V}}_{\mathrm{i}}^{\mathrm{\gamma }-1} \\ {\mathrm{T}}_{\mathrm{f}} = {\mathrm{T}}_{\mathrm{i} }{4}^{1.5-1} = 273 \times \sqrt{4} = 546 \mathrm{K} \\ \mathrm{Now} \mathrm{the} \mathrm{difference}, ∆\mathrm{T} = {\mathrm{T}}_{\mathrm{f}} - {\mathrm{T}}_{\mathrm{i}} = 546 - 273 = 273 \mathrm{K} \end{gathered}$$

$$\Delta S = \int_{T_1}^{T_2} \frac{dQ}{T}$$​

For a system where the temperature is slowly raised and the specific heat capacity is constant, the heat supplied ($dQ$) is:

$$dQ = m s \, dT$$

Substituting into the entropy formula:

$$\Delta S = \int_{T_1}^{T_2} \frac{ms \, dT}{T}$$​

This simplifies to:

$$\Delta S = ms \int_{T_1}^{T_2} \frac{dT}{T}$$​

The integral of $\frac{1}{T}$​ is $\ln T$, so:

$$\Delta S = ms \left[\ln T\right]_{T_1}^{T_2}$$

​​ $$\Delta S = ms \left(\ln T_2 - \ln T_1\right)$$

Using logarithmic properties:

$$\Delta S = ms \ln \frac{T_2}{T_1}$$​​

Thus, the change in entropy is:

$$\boxed{ms \ln \frac{T_2}{T_1}}$$​​​

(A) In adiabatic process
$T{V}^{\gamma -1}=$ constant
So, temperature increases if volume decreases (R) Free expansion occurs fast, so Adiabatic

$$\begin{gathered} I= dQ \\ dQ =TdS \\ dS =\frac{dQ}{T} \\ \int dS ={\int }_{T_1}^{T}mC\frac{dT}{T} \\ \Delta S =mC[\ln T{]}_{T_1}^{T_2} \end{gathered}$$

$\Delta S=mc\ln \frac{T_2}{T_1}$

$$\begin{gathered} \text{ Efficiency }\left(\eta \right)=1-\frac{T_2}{T_1} \\ {\eta }_{12}=1-\frac{273}{473}=\frac{200}{473} \\ {\eta }_1=1-\frac{373}{473}=\frac{100}{473} \\ {\eta }_2=1-\frac{273}{373} \\ =\frac{100}{373} \\ {\eta }_1+{\eta }_2>\frac{200}{473} \\ {\eta }_1+{\eta }_2>{\eta }_{12} \end{gathered}$$

1. Condition:

   • If pressure $P \propto T$, this implies: $$P = kT \quad \text{(where \( k \) is constant)}.$$
   • This relationship is only valid if the volume $V$ is constant, as per the ideal gas law:
   • $$PV = nRT \quad \implies \quad P \propto T \quad \text{(if \( V \) is constant)}.$$.
   • Thus, the process is isochoric (the statement $A$ is correct).

2. Work Done:

   • In an isochoric process, the volume remains constant ($\Delta V = 0$), so the work done: $$W = P \Delta V = 0.$$ Hence, $B$ is also correct.

3. Internal Energy:

   • The internal energy $U$ of an ideal gas depends on temperature: $$\Delta U = n C_v \Delta T.$$. Since temperature increases, internal energy also increases. Thus, $C$ is correct.

4. Conclusion:

   • All statements $A$, $B$, and $C$ are correct.

Answer: (3) $A, B,$ and $C$ are correct.

Given: \(P \propto T^3\)
In adiabatic process
\(T^{\prime} P^{1-\gamma}=\) constant
\(
\begin{aligned}
& T \alpha \frac{1}{p^{(1-\gamma) / \gamma}} \\
& T^{(\gamma /-1)} \propto P
\end{aligned}
\)

Comparing equations (i) and (ii), we get
\(
\begin{aligned}
& \therefore \quad \frac{\gamma}{\gamma-1}=3 \\
& 3 \gamma-3=\gamma \\
& 2 \gamma=3 \\
& \frac{C_P}{C_P}=\gamma=\frac{3}{2}
\end{aligned}
\)

Extensive variables depend on the system size (e.g., mass, volume, and internal energy). Intensive variables do not depend on the system size (e.g., pressure, temperature, and density).

• Assertion: Correct

  • In an isothermal process ($PV =$ constant), an increase in pressure leads to a larger decrease in volume compared to an adiabatic process ($PV^\gamma =$constant) because temperature remains constant in the isothermal case.

• Reason: Correct

  • The adiabatic process follows the equation $PV^\gamma =$ constant, which correctly describes its behavior.

• Logical Connection:

  • Since the reason correctly explains the assertion, both are correct.

$$\boxed{B} \quad \text{Assertion is correct and Reason is correct.}$$B​Assertion is correct and Reason is correct.

From the first law of thermodynamics for an isobaric process: $$\Delta Q = \Delta U + \Delta W.$$

• For a monoatomic gas:
  • $\Delta U = \frac{3}{2} n R \Delta T$,
  • $\Delta W = n R \Delta T$.
• Heat supplied: $$\Delta Q = \frac{3}{2} n R \Delta T + n R \Delta T = \frac{5}{2} n R \Delta T.$$
• Ratio: $$\frac{\Delta Q}{\Delta U} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3}.$$

Thus, the ratio is $\frac{5}{3}$.

Work done in adiabatic process for ideal gas is equal to change in internal energy.

The volume decreases more rapidly in an adiabatic process compared to an isothermal process. This is because in the isothermal process, heat exchange allows the gas to expand or compress without changing its temperature, whereas in the adiabatic process, the gas is compressed without heat exchange, causing a sharper decrease in volume.
Reason (R) is true, as it correctly describes the relationship between pressure and volume in both processes.

From the equation, we get:

$\frac{P}{T}=\text{ constant }$

Now, let's break down each statement:
A. The work done by the gas during the process is zero.

Since the volume remains constant, there is no displacement of the gas, and hence no work is done. So, this statement is true.
B. The heat added to the gas is different from the change in its internal energy.
This is incorrect. For an isochoric process, the heat added to the system equals the change in the internal energy, as there is no work done. So, this statement is false.
C. The volume of the gas is increased. Since the process is isochoric (constant volume), this statement is false.
D. The internal energy of the gas is increased.

Since the temperature increases in this process, and for an ideal gas, the internal energy depends only on temperature, the internal energy increases. So, this statement is true.
E. The process is isochoric (constant volume process).
As derived from the relationship P / T = constant, the volume remains constant, making this process isochoric. So, this statement is true.
The correct answer is A, D, E Only.

$$\begin{gathered} I= dQ \\ dQ =TdS \\ dS =\frac{dQ}{T} \\ \int dS ={\int }_{T_1}^{T}mC\frac{dT}{T} \\ \Delta S =mC[\ln T{]}_{T_1}^{T_2} \end{gathered}$$

$\Delta S=mc\ln \frac{T_2}{T_1}$

• Assertion: Correct

  • In an isothermal process ($PV =$ constant), an increase in pressure leads to a larger decrease in volume compared to an adiabatic process ($PV^\gamma =$constant) because temperature remains constant in the isothermal case.

• Reason: Correct

  • The adiabatic process follows the equation $PV^\gamma =$ constant, which correctly describes its behavior.

• Logical Connection:

  • Since the reason correctly explains the assertion, both are correct.

$$\boxed{B} \quad \text{Assertion is correct and Reason is correct.}$$B​Assertion is correct and Reason is correct.