Adiabatic Process Questions in English

(C) For an adiabatic process,the heat exchange $Q = 0$.
According to the first law of thermodynamics,$Q = \Delta U + W$. Since $Q = 0$,we have $W = -\Delta U$. This means the work done is equal to the negative change in internal energy,or $|W| = |\Delta U|$. Thus,statement $(B)$ is correct.
For an ideal gas,the internal energy change is $\Delta U = nC_v \Delta T$. Therefore,the work done $W = -nC_v(T_2 - T_1) = nC_v(T_1 - T_2)$. The magnitude of work done is proportional to $(T_2 - T_1)$. Thus,statement $(E)$ is correct.
Statements $(A), (C),$ and $(D)$ are incorrect because they describe isothermal or other processes,not adiabatic processes.
Therefore,the correct option is $(B), (E)$ only.

(D) Given: Initial volume $V_1 = 800 \ cm^3$,Final volume $V_2 = 200 \ cm^3$,Initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$,and adiabatic index $\gamma = 1.5$.
For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values: $300 \times (800)^{1.5-1} = T_2 \times (200)^{1.5-1}$.
$300 \times (800)^{0.5} = T_2 \times (200)^{0.5}$.
$T_2 = 300 \times \left( \frac{800}{200} \right)^{0.5} = 300 \times (4)^{0.5} = 300 \times 2 = 600 \ K$.
The change in temperature is $\Delta T = T_2 - T_1 = 600 \ K - 300 \ K = 300 \ K$.

(C) Since the gas is in a thermally insulated container and is compressed suddenly,the process is adiabatic.
For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P_i V_i^\gamma = P_f V_f^\gamma$.
Here,$V_f = \frac{1}{8} V_i$,which implies $\frac{V_i}{V_f} = 8$.
The ratio of final pressure to initial pressure is $\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^\gamma$.
Substituting the values,$\frac{P_f}{P_i} = (8)^{5/3}$.
Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$.
Thus,the ratio of final pressure to initial pressure is $32$.

(B) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Given the relation $TV^{x} = \text{constant}$, we compare the exponents of $V$.
Thus, $x = \gamma - 1$.
For a monoatomic ideal gas, the adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.
Substituting the value of $\gamma$ into the expression for $x$:
$x = \frac{5}{3} - 1 = \frac{2}{3}$.

(D) Given: $\mu = 1 \text{ mole}$,$\gamma = \frac{5}{3}$,$T_1 = 127^{\circ} C = 400 \ K$,$V_2 = \frac{8}{27} V_1$.
For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = 400 \times \left( \frac{27}{8} \right)^{\frac{5}{3}-1} = 400 \times \left( \frac{27}{8} \right)^{\frac{2}{3}} = 400 \times \left( \left( \frac{3}{2} \right)^3 \right)^{\frac{2}{3}} = 400 \times \left( \frac{3}{2} \right)^2 = 400 \times \frac{9}{4} = 900 \ K$.
Work done on the system is $W = -\frac{\mu R (T_2 - T_1)}{\gamma - 1}$.
Using $R = 2 \text{ cal/mol K}$,$W = -\frac{1 \times 2 \times (900 - 400)}{\frac{5}{3} - 1} = -\frac{2 \times 500}{\frac{2}{3}} = -\frac{1000 \times 3}{2} = -1500 \text{ cal}$.
The magnitude of work done on the system is $1500 \text{ cal}$.

(A) Given that,$V_2 = \frac{V_1}{4}$ and $\gamma = 1.5$.
Since the compression is sudden,the process is adiabatic.
For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values,we get $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$.
Since $\frac{V_1}{V_2} = 4$ and $\gamma - 1 = 1.5 - 1 = 0.5$,we have $T_2 = T_1 (4)^{0.5} = T_1 \times 2 = 2 T_1$.
The increase in temperature is $\Delta T = T_2 - T_1 = 2 T_1 - T_1 = T_1$.
Given that the gas is at normal temperature,$T_1 = 273 \ K$.
Therefore,the increase in temperature is $273 \ K$.

(C) For an adiabatic compression,the relation is $PV^{\gamma} = \text{constant}$.
Given: $V_{\text{new}} = \frac{1}{4} V$ and $\gamma = \frac{3}{2}$.
Using the adiabatic process equation:
$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$
$P \cdot V^{\gamma} = P_{\text{new}} \cdot \left(\frac{V}{4}\right)^{\gamma}$
$\frac{P_{\text{new}}}{P} = \left(\frac{V}{V/4}\right)^{\gamma} = (4)^{\gamma}$
$\frac{P_{\text{new}}}{P} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$
Therefore,$P_{\text{new}} = 8P$.

(A) The first law of thermodynamics is given by $Q = \Delta U + W$,where $Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $W$ is the work done by the system.
For an adiabatic process,there is no exchange of heat with the surroundings,so $Q = 0$.
Substituting $Q = 0$ into the first law equation: $0 = \Delta U + W$.
Rearranging this gives $\Delta U = -W$.
Therefore,the statement $\Delta U = -W$ represents an adiabatic process.

(C) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by Poisson's equation: $p V^{\gamma} = \text{constant}$.
Using the ideal gas law $pV = RT$,we can write $V = \frac{RT}{p}$.
Substituting this into the adiabatic equation:
$p \left( \frac{RT}{p} \right)^{\gamma} = \text{constant}$
$p^{1-\gamma} T^{\gamma} = \text{constant}$
$p \propto T^{\frac{\gamma}{\gamma-1}}$.
Comparing this with the given relation $p \propto T^{C}$,we get $C = \frac{\gamma}{\gamma-1}$.
For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
Substituting the value of $\gamma$:
$C = \frac{5/3}{5/3 - 1} = \frac{5/3}{2/3} = \frac{5}{2}$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
In an adiabatic process,the heat exchange $\Delta Q = 0$.
Therefore,$0 = \Delta U + \Delta W$,which implies $\Delta U = -\Delta W$.
When a gas is compressed,work is done on the gas,so $\Delta W$ is negative.
Substituting this into the equation,$\Delta U = -(\text{negative value})$,which results in a positive $\Delta U$.
$A$ positive change in internal energy $(\Delta U > 0)$ means the internal energy of the gas increases.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta Q = \Delta U + W$,where $\Delta Q$ is the heat supplied to the system and $W$ is the work done by the system.
For an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + W$.
Therefore,$\Delta U = -W$.

(D) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $T_1 = T$,$V_1 = V$,$V_2 = 2V$,and $\gamma = 3/2$.
Substituting these values: $T V^{\gamma-1} = T_2 (2V)^{\gamma-1}$.
$T_2 = T \left(\frac{V}{2V}\right)^{\gamma-1} = T \left(\frac{1}{2}\right)^{3/2-1} = T \left(\frac{1}{2}\right)^{1/2} = \frac{T}{\sqrt{2}}$.
The work done in an adiabatic process is $W = \frac{R(T_1 - T_2)}{\gamma - 1}$.
Substituting the values: $W = \frac{R(T - T/\sqrt{2})}{3/2 - 1} = \frac{R T (1 - 1/\sqrt{2})}{1/2}$.
$W = 2 R T \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right) = \sqrt{2} R T (\sqrt{2}-1) = R T (2 - \sqrt{2})$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $TV^{\gamma-1} = \text{constant}$.
Thus,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Given $V_i = V_0$,$V_f = \frac{V_0}{32}$,and $\gamma = \frac{7}{5}$.
Then $\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$.
Substituting the values: $T_i (V_0)^{2/5} = T_f \left(\frac{V_0}{32}\right)^{2/5}$.
$T_f = T_i \left(\frac{V_0}{V_0/32}\right)^{2/5} = T_i (32)^{2/5}$.
Since $32 = 2^5$,we have $T_f = T_i (2^5)^{2/5} = T_i (2^2) = 4T_i$.
Comparing $T_f = xT_i$ with $T_f = 4T_i$,we get $x = 4$.

(B) For an adiabatic process,the relation between pressure $(P)$ and temperature $(T)$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$.
This can be rewritten as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Comparing this with $P \propto T^{c}$,we get $c = \frac{\gamma}{\gamma-1}$.
For a rigid diatomic gas,the degrees of freedom $(f)$ is $5$.
The adiabatic exponent $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1.4$.
Substituting the value of $\gamma$ into the expression for $c$:
$c = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = \frac{14}{4} = 3.5$.
Therefore,the value of $c$ is $3.5$.

(D) The molar specific heat $C$ for a polytropic process $PV^{n} = \text{constant}$ is given by the formula: $C = C_{V} + \frac{R}{1-n}$.
Given that the specific heat $C = 0$, we have: $0 = C_{V} + \frac{R}{1-n}$.
Substituting $C_{V} = \frac{R}{\gamma-1}$, we get: $0 = \frac{R}{\gamma-1} + \frac{R}{1-n}$.
Rearranging the terms: $\frac{R}{n-1} = \frac{R}{\gamma-1}$.
This implies $n-1 = \gamma-1$, which simplifies to $n = \gamma$.
Therefore, the process is an adiabatic process, and the value of $n$ is $\gamma$.

(A) For an adiabatic process, the relationship between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.
Since the gas is monoatomic, the adiabatic index $\gamma = 5/3$.
Therefore, $\gamma - 1 = 5/3 - 1 = 2/3$.
The volume of the gas in a cylinder of cross-sectional area $A$ is $V = A \times L$.
Thus, $T_1 (A L_1)^{2/3} = T_2 (A L_2)^{2/3}$.
Rearranging the terms, we get $(T_2 / T_1) = (L_1 / L_2)^{2/3}$.

(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Let the initial pressure be $P_1 = P_0$ and initial volume be $V_1 = V$.
The final volume is $V_2 = \frac{V}{8}$.
The adiabatic index is given as $\gamma = \frac{4}{3}$.
Using the relation $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$,we have:
$P_0 V^{\gamma} = P_2 \left(\frac{V}{8}\right)^{\gamma}$.
$P_2 = P_0 \left(\frac{V}{V/8}\right)^{\gamma} = P_0 (8)^{\gamma}$.
Substituting $\gamma = \frac{4}{3}$:
$P_2 = P_0 (8)^{4/3} = P_0 (2^3)^{4/3} = P_0 (2^4) = 16 P_0$.
Thus,the new pressure is $16 P_0$.

(B) According to the first law of thermodynamics, the change in internal energy $(\Delta U)$ is given by $\Delta U = Q + W$, where $Q$ is the heat added to the system and $W$ is the work done on the system.
Given that the increase in internal energy is equal to the work done on the system, we have $\Delta U = W$.
Comparing this with the first law equation, we get $Q = 0$.
A thermodynamic process in which no heat is exchanged with the surroundings $(Q = 0)$ is known as an adiabatic process.
Therefore, the correct option is $B$.

(B) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
For a monoatomic ideal gas, the adiabatic index $\gamma = \frac{5}{3}$.
Therefore, $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Given the initial state $(T_1, V_1) = (T, V)$ and the final state $(T_2, V_2) = (xT, V/27)$.
Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
$T(V)^{2/3} = (xT) \left(\frac{V}{27}\right)^{2/3}$.
$1 = x \left(\frac{1}{27}\right)^{2/3}$.
$1 = x \left(\left(\frac{1}{3^3}\right)^{1/3}\right)^2 = x \left(\frac{1}{3^2}\right) = x \left(\frac{1}{9}\right)$.
Thus, $x = 9$.

(A) The r.m.s. velocity of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$, reducing the $v_{rms}$ by a factor of $4$ means the temperature $T$ must be reduced by a factor of $4^2 = 16$.
So, $T_f = \frac{T_i}{16}$.
For an adiabatic process, the relationship between temperature and volume is $TV^{\gamma-1} = \text{constant}$.
Thus, $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
Substituting the values, $T_i V_i^{1.5-1} = \frac{T_i}{16} V_f^{1.5-1}$.
$V_i^{0.5} = \frac{1}{16} V_f^{0.5}$.
$\sqrt{V_i} = \frac{1}{16} \sqrt{V_f}$.
$\sqrt{\frac{V_f}{V_i}} = 16$.
$\frac{V_f}{V_i} = 16^2 = 256$.
Therefore, the gas must be expanded $256$ times.

(C) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Thus,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given that the gas is monoatomic,the adiabatic index $\gamma = 5/3$.
Therefore,$\gamma - 1 = 5/3 - 1 = 2/3$.
The final volume $V_2 = 12.5 \% \text{ of } V_1 = 0.125 V_1 = (1/8) V_1$.
Substituting the values into the equation: $T_1 V_1^{2/3} = T_2 (V_1/8)^{2/3}$.
$T_2 = T_1 \times (V_1 / (V_1/8))^{2/3} = T_1 \times (8)^{2/3}$.
$T_2 = T_1 \times (2^3)^{2/3} = T_1 \times 2^2 = 4T_1$.
Comparing this with $T_2 = xT_1$,we get $x = 4$.

(D) For a sudden compression,the process is adiabatic.
In an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: Initial volume $V_1 = 360 \text{ c.c.}$,final volume $V_2 = 90 \text{ c.c.}$,initial pressure $P_1 = P$,and adiabatic index $\gamma = 5/2$.
Substituting the values into the equation:
$P \times (360)^{5/2} = P_2 \times (90)^{5/2}$
$P_2 = P \times \left( \frac{360}{90} \right)^{5/2}$
$P_2 = P \times (4)^{5/2}$
$P_2 = P \times (2^2)^{5/2}$
$P_2 = P \times 2^5$
$P_2 = 32P$
Therefore,the final pressure is $32P$.

(D) For an adiabatic process,the work done by the gas is given by the formula:
$W = \frac{nR(T_i - T_f)}{\gamma - 1}$
Given:
Number of moles $n = 1$
Work done $W = 6R$
Initial temperature $T_i = T$
Ratio of specific heats $\gamma = 5/3$
Substituting these values into the formula:
$6R = \frac{1 \cdot R(T - T_f)}{(5/3) - 1}$
$6R = \frac{R(T - T_f)}{2/3}$
$6R = \frac{3R(T - T_f)}{2}$
Dividing both sides by $R$:
$6 = \frac{3(T - T_f)}{2}$
$12 = 3(T - T_f)$
$4 = T - T_f$
$T_f = T - 4$
Therefore,the final temperature of the gas is $(T - 4) \ K$.

(C) In a $P-V$ diagram,the slope of an adiabatic process is given by $\frac{dP}{dV} = -\gamma \frac{P}{V}$,while the slope of an isothermal process is given by $\frac{dP}{dV} = -\frac{P}{V}$.
Since the adiabatic index $\gamma > 1$,the adiabatic curve is steeper than the isothermal curve.
Looking at the provided $P-V$ graph,the segments $AD$ and $BC$ exhibit a steeper slope compared to the segments $AB$ and $CD$.
Therefore,the adiabatic processes are represented by the regions $AD$ and $BC$.

(A) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $T^{\gamma} P^{1-\gamma} = \text{constant}$,which can be rewritten as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
For a diatomic gas,the adiabatic index $\gamma = \frac{7}{5} = 1.4$.
Substituting the value of $\gamma$ into the exponent $x = \frac{\gamma}{\gamma-1}$:
$x = \frac{1.4}{1.4-1} = \frac{1.4}{0.4} = \frac{14}{4} = 3.5$.
Thus,$P \propto T^{3.5}$,so the value of $x$ is $3.5$.

(A) For an adiabatic process,the relation between pressure $P$ and temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$,which can be rewritten as $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Given $P \propto T^K$,we have $K = \frac{\gamma}{\gamma-1}$.
We know that $C_p = C_v + R$. Given $R = 0.4 C_v$,we have $C_p = C_v + 0.4 C_v = 1.4 C_v$.
The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{1.4 C_v}{C_v} = 1.4 = \frac{14}{10} = \frac{7}{5}$.
Substituting $\gamma = 7/5$ into the expression for $K$:
$K = \frac{7/5}{(7/5) - 1} = \frac{7/5}{2/5} = \frac{7}{2}$.

(A) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
For a monoatomic ideal gas,the adiabatic exponent is $\gamma = \frac{5}{3}$.
Thus,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Since the gas is in a cylinder of constant cross-sectional area $A$,the volume is $V = A \times L$. Therefore,$V_1 = A L_1$ and $V_2 = A L_2$.
Substituting these into the temperature ratio equation:
$\frac{T_2}{T_1} = \left(\frac{A L_1}{A L_2}\right)^{2/3} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given: $T_1 = 27^{\circ} C = 300 \ K$,$V_2 = (8/27) V_1$,and $\gamma = 5/3$.
Substituting the values:
$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1} = \left(\frac{V_1}{(8/27)V_1}\right)^{(5/3)-1} = \left(\frac{27}{8}\right)^{2/3}$.
$\frac{T_2}{T_1} = \left(\left(\frac{3}{2}\right)^3\right)^{2/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25$.
$T_2 = 2.25 \times 300 \ K = 675 \ K$.
The rise in temperature is $\Delta T = T_2 - T_1 = 675 \ K - 300 \ K = 375 \ K$.

(C) Since the compression is sudden,the process is adiabatic.
For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: $V_2 = \frac{V_1}{4}$,so $\frac{V_1}{V_2} = 4$.
Given: $\gamma = 1.5 = \frac{3}{2}$.
Substituting these values into the equation:
$\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma = (4)^{3/2}$.
Calculating the value: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,the final pressure $P_2 = 8 P_1$,which is $8$ times the initial pressure.

(A) In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat exchange occurs $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$. Since $Q = 0$,we have $\Delta U = -W$,which means all work done is at the expense of internal energy.
The equation of state for an adiabatic process is $PV^{\gamma} = \text{constant}$,where $\gamma$ is the adiabatic index. The statement $PV = \text{constant}$ describes an isothermal process,not an adiabatic one.
Therefore,the statement '$PV=$ constant' is incorrect.

(B) According to the first law of thermodynamics,the heat supplied to the system $\Delta Q$ is given by $\Delta Q = \Delta U + dW$.
For an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + dW$.
Rearranging the terms,we get $\Delta U = -dW$.
Therefore,the condition $\Delta U = -dW$ corresponds to an adiabatic process.

(B) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = \text{constant}$.
Since density $\rho = \frac{m}{V}$,we have $V = \frac{m}{\rho}$. Substituting this into the adiabatic equation:
$P \left(\frac{m}{\rho}\right)^{\gamma} = \text{constant}$
$P \rho^{-\gamma} = \text{constant}$
Therefore,$\frac{P^{\prime}}{P} = \left(\frac{\rho^{\prime}}{\rho}\right)^{\gamma}$.
Given $\frac{\rho^{\prime}}{\rho} = 32$ and $\gamma = \frac{7}{5}$,we substitute these values:
$\frac{P^{\prime}}{P} = (32)^{\frac{7}{5}}$
$\frac{P^{\prime}}{P} = (2^5)^{\frac{7}{5}} = 2^7 = 128$.

(B) The correct option is $B$.
Concept: In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat is transferred into or out of the system $(dQ = 0)$.
Reason: This process typically occurs rapidly,preventing sufficient time for heat exchange to take place between the system and the surroundings.

(C) The change in internal energy $(\Delta U)$ of an ideal gas depends only on the change in temperature and is given by the formula: $\Delta U = nC_v\Delta T$.
For an ideal gas, the molar heat capacity at constant volume is $C_v = \frac{R}{\gamma-1}$.
Substituting this into the internal energy formula, we get: $\Delta U = n \left(\frac{R}{\gamma-1}\right) (T_2 - T_1)$.
Therefore, the change in internal energy is $\frac{nR}{\gamma-1}(T_2 - T_1)$.

(D) For a sudden compression,the process is adiabatic.
For an adiabatic process,the relation between pressure and volume is $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Given: $V_1 = V$,$V_2 = \frac{V}{4}$,$P_1 = P$,and $\gamma = \frac{5}{2}$.
Substituting these values into the adiabatic equation:
$P \cdot V^{\gamma} = P_2 \cdot \left(\frac{V}{4}\right)^{\gamma}$
$P_2 = P \cdot \left(\frac{V}{V/4}\right)^{\gamma}$
$P_2 = P \cdot (4)^{\gamma}$
$P_2 = P \cdot (4)^{5/2}$
$P_2 = P \cdot (2^2)^{5/2}$
$P_2 = P \cdot 2^5$
$P_2 = 32 P$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta Q = \Delta U + W$,where $\Delta Q$ is the heat supplied to the system and $W$ is the work done by the system.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + W$.
Therefore,$\Delta U = -W$.

(B) For an adiabatic process,the relationship between pressure $p$ and volume $V$ is given by $p V^{\gamma} = \text{constant}$.
Let the initial pressure be $p$ and the initial volume be $V$.
The final volume is $V' = \frac{V}{8}$.
Using the adiabatic equation: $p V^{\gamma} = P' (V')^{\gamma}$.
Substituting the given values: $p V^{4/3} = P' \left(\frac{V}{8}\right)^{4/3}$.
$P' = p \left(\frac{V}{V/8}\right)^{4/3} = p (8)^{4/3}$.
Since $8 = 2^3$,we have $P' = p (2^3)^{4/3} = p (2^4) = 16p$.
Therefore,the new pressure is $16p$.

(B) For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = \text{constant}$.
For the adiabatic path $BC$,the points $B$ and $C$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:
$T_1 V_b^{\gamma-1} = T_2 V_c^{\gamma-1}$
$\Rightarrow \left(\frac{V_b}{V_c}\right)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(1)$
For the adiabatic path $AD$,the points $A$ and $D$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:
$T_1 V_a^{\gamma-1} = T_2 V_d^{\gamma-1}$
$\Rightarrow \left(\frac{V_a}{V_d}\right)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(2)$
Equating $(1)$ and $(2)$:
$\left(\frac{V_b}{V_c}\right)^{\gamma-1} = \left(\frac{V_a}{V_d}\right)^{\gamma-1}$
$\Rightarrow \frac{V_b}{V_c} = \frac{V_a}{V_d}$
Rearranging the terms,we get:
$\frac{V_b}{V_a} = \frac{V_c}{V_d}$
Therefore,option $(B)$ is correct.

(C) Since the compression is sudden,it is an adiabatic process.
For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given: $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$,$V_1 = V$,$V_2 = V/8$,and $\gamma = 5/3$.
Substituting the values:
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$
$T_2 = 300 \times \left( \frac{V}{V/8} \right)^{(5/3) - 1}$
$T_2 = 300 \times (8)^{2/3}$
$T_2 = 300 \times (2^3)^{2/3} = 300 \times 2^2 = 300 \times 4 = 1200 \ K$.

(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by the equation $T V^{\gamma-1} = \text{constant}$.
Therefore,for two different states $(P_1, V_1, T_1)$ and $(P_2, V_2, T_2)$,the relation holds as $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Thus,the correct option is $A$.

(C) For an adiabatic process, the relation between pressure and volume is given as $PV^{\gamma} = \text{constant}$. Here, $\gamma = 3/2$.
Using the ideal gas law $PV = nRT$, we can write $P = nRT/V$.
Substituting this into the adiabatic equation: $(nRT/V)V^{\gamma} = \text{constant}$, which simplifies to $TV^{\gamma-1} = \text{constant}$.
Given $\gamma = 3/2$, the relation becomes $TV^{(3/2 - 1)} = TV^{1/2} = \text{constant}$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given $T_1 = T$ and $V_2 = V_1/2$.
Using $T_1 V_1^{1/2} = T_2 V_2^{1/2}$:
$T_2 = T_1 (V_1/V_2)^{1/2} = T (V_1 / (V_1/2))^{1/2} = T (2)^{1/2} = \sqrt{2}T$.

(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Given the condition $T \propto \frac{1}{\sqrt{V}}$,we can write this as $T \propto V^{-1/2}$,which implies $TV^{1/2} = \text{constant}$.
Comparing the two expressions $TV^{\gamma-1} = \text{constant}$ and $TV^{1/2} = \text{constant}$,we get $\gamma - 1 = \frac{1}{2}$.
Solving for $\gamma$,we find $\gamma = 1 + 0.5 = 1.5$.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta Q = \Delta U + dW$,where $\Delta Q$ is the heat supplied to the system and $dW$ is the work done by the system.
In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the equation,we get $0 = \Delta U + dW$.
Therefore,$\Delta U = -dW$.

(D) For an adiabatic process,the relationship between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given that the gas is compressed to $(1/8)^{\text{th}}$ of its initial volume,we have $V_2 = V_1 / 8$,which implies $V_1 / V_2 = 8$.
The adiabatic index is given as $\gamma = 4/3$.
Substituting these values into the adiabatic equation:
$P_2 / P_1 = (V_1 / V_2)^\gamma$
$P_2 / P_0 = (8)^{4/3}$
$P_2 / P_0 = (2^3)^{4/3} = 2^4 = 16$
Therefore,the new pressure is $P_2 = 16 P_0$.

(D) For an adiabatic process,the relation between temperature and volume is $TV^{\gamma-1} = \text{constant}$.
Thus,$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
Given $T_1 = 27^{\circ} C = 300 \ K$ and $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.
Given $\gamma = 5/3$,so $\gamma - 1 = 5/3 - 1 = 2/3$.
Substituting the values: $\frac{T_2}{300} = \left(\frac{27}{8}\right)^{2/3} = \left(\left(\frac{3}{2}\right)^3\right)^{2/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
$T_2 = \frac{9}{4} \times 300 = 9 \times 75 = 675 \ K$.
The rise in temperature is $\Delta T = T_2 - T_1 = 675 \ K - 300 \ K = 375 \ K$.

(D) For an ideal gas,density $\rho = \frac{m}{V}$. Since the mass $m$ remains constant,$\rho \propto \frac{1}{V}$.
Given that the final density is twice the initial density,$\rho_2 = 2\rho_1$,which implies $V_2 = \frac{V_1}{2}$ or $\frac{V_1}{V_2} = 2$.
For an adiabatic process,the relation between pressure and volume is $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Therefore,the final pressure $P_2$ is given by $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma$.
Substituting the given values,$P_2 = p \times (2)^{7/5} = p \times (2)^{1.4}$.
Calculating the value,$2^{1.4} \approx 2.639$.
Thus,the final pressure is approximately $2.63p$.

(A) Given: Initial volume $V_1$,Final volume $V_2 = V_1/8$,Adiabatic index $\gamma = 5/3$.
For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Rearranging for the ratio of final pressure to initial pressure: $\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma$.
Substituting the values: $\frac{P_2}{P_1} = (8)^{5/3}$.
Since $8 = 2^3$,we have $(2^3)^{5/3} = 2^5 = 32$.
Thus,the ratio of the final pressure to the initial pressure is $32$.

(C) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
From the ideal gas equation, $PV = RT$, we have $V = \frac{RT}{P}$.
Substituting $V$ in the adiabatic equation:
$P \left( \frac{RT}{P} \right)^{\gamma} = \text{constant}$
$P \cdot \frac{T^{\gamma}}{P^{\gamma}} = \text{constant}'$
$P^{1-\gamma} T^{\gamma} = \text{constant}'$
$P^{1-\gamma} = \frac{\text{constant}'}{T^{\gamma}}$
$P = \text{constant}'' \cdot T^{\frac{\gamma}{\gamma-1}}$
Comparing this with $P \propto T^{x}$, we get $x = \frac{\gamma}{\gamma-1}$.
For a diatomic gas, the adiabatic exponent $\gamma = 1.4$ (or $\frac{7}{5}$).
Substituting the value of $\gamma$:
$x = \frac{1.4}{1.4 - 1} = \frac{1.4}{0.4} = 3.5$.
Therefore, the correct option is $C$.