Adiabatic Process Questions in English

(B) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.
Given: $n = 2 \ mol$,$T_1 = 27^{\circ}C = 300 \ K$,$V_1 = V$,$V_2 = 2V$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$,so $\gamma - 1 = 2/3$.
$(a)$ Substituting the values: $300 \times V^{2/3} = T_2 \times (2V)^{2/3}$.
$T_2 = 300 / (2^{2/3}) \approx 300 / 1.5874 \approx 189 \ K$.
$(b)$ The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a monoatomic gas,$C_v = (3/2)R$.
$\Delta U = 2 \times (3/2) \times 8.314 \times (189 - 300) \approx 3 \times 8.314 \times (-111) \approx -2768 \ J \approx -2.7 \ kJ$.

(D) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by $p V^{\gamma} = \text{constant}$.
Since density $\rho$ is defined as $\rho = \frac{m}{V}$,we have $V = \frac{m}{\rho}$.
Substituting this into the adiabatic equation:
$p \left( \frac{m}{\rho} \right)^{\gamma} = \text{constant}$.
Since mass $m$ is constant,$m^{\gamma}$ is also a constant.
Therefore,$p \cdot \frac{1}{\rho^{\gamma}} = \text{constant}$.
This can be written as $p \rho^{-\gamma} = \text{constant}$.

(B) For a cyclic process $ABCA$,the change in internal energy is $\Delta U_{ABCA} = 0$.
From the first law of thermodynamics,$Q_{ABCA} = W_{ABCA}$.
This can be written as: $Q_{A \to B} + Q_{B \to C} + Q_{C \to A} = W_{A \to B} + W_{B \to C} + W_{C \to A} \quad (1)$.
Given that the heat interaction along path $B \to C \to A$ is equal to the work done along path $A \to B \to C$,we have:
$Q_{B \to C} + Q_{C \to A} = W_{A \to B} + W_{B \to C} \quad (2)$.
Subtracting equation $(2)$ from equation $(1)$:
$Q_{A \to B} = W_{C \to A}$.
From the $P-V$ diagram,the process $C \to A$ is an isochoric process (constant volume),so the work done $W_{C \to A} = 0$.
Therefore,$Q_{A \to B} = 0$.
$A$ process in which the heat interaction is zero is an adiabatic process. Thus,the process $A \to B$ is adiabatic.

(C) Given the process law: $T P^{-2/5} = \text{constant}$.
Using the ideal gas equation $PV = nRT$, we have $P = \frac{nRT}{V}$.
Substituting $P$ into the given equation: $T (\frac{nRT}{V})^{-2/5} = \text{constant}$.
This simplifies to $T \cdot T^{-2/5} \cdot V^{2/5} = \text{constant}$, which gives $T^{3/5} V^{2/5} = \text{constant}$.
Raising both sides to the power of $5/3$, we get $T V^{2/3} = \text{constant}$.
Since $T \propto PV$, we have $(PV) V^{2/3} = \text{constant}$, which implies $P V^{5/3} = \text{constant}$.
This is an adiabatic process with adiabatic index $\gamma = 5/3$ (for helium, a monoatomic gas).
For an adiabatic process, the heat exchanged $Q = 0$.

(A) For an ideal gas,the equation of state is given by $PV = nRT$.
In an adiabatic expansion,the system does work on the surroundings at the expense of its internal energy.
Since the internal energy of an ideal gas depends only on temperature $(U \propto T)$,a decrease in internal energy leads to a decrease in temperature $(T)$.
As $n$ and $R$ are constants,if the temperature $(T)$ decreases during the adiabatic expansion,the product $PV$ must also decrease.

(A) For an adiabatic process, the equation is $PV^{\gamma} = \text{constant}$.
Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} + P \gamma V^{\gamma-1} = 0$.
Thus, the slope $m = \frac{dP}{dV} = -\gamma \frac{P}{V}$.
From the figure, the magnitude of the slope of curve $B$ is greater than the magnitude of the slope of curve $A$ $(|m_B| > |m_A|)$.
Therefore, $\gamma_B > \gamma_A$.
For monoatomic gases (like $He$, $Ar$), $\gamma = 1.67$.
For diatomic gases (like $O_2$), $\gamma = 1.4$.
Since $\gamma_B > \gamma_A$, curve $B$ represents a monoatomic gas and curve $A$ represents a diatomic gas.
Thus, $A$ corresponds to $O_2$ and $B$ corresponds to $He$ (or $Ar$).
Comparing with the options, $A$ is $O_2$ and $B$ is $He$ fits the condition.

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = \text{constant}$.
Since density $d = \frac{m}{V},$ we have $V = \frac{m}{d},$ which implies $V \propto \frac{1}{d}.$
Substituting this into the adiabatic equation: $P \left(\frac{1}{d}\right)^{\gamma} = \text{constant},$ or $P \propto d^{\gamma}.$
Therefore,$\frac{P'}{P} = \left(\frac{d'}{d}\right)^{\gamma}.$
Given $\gamma = 7/5$ and $\frac{d'}{d} = 32,$
$\frac{P'}{P} = (32)^{7/5} = (2^5)^{7/5} = 2^{5 \times (7/5)} = 2^7.$
Calculating the value,$2^7 = 128.$
Thus,$\frac{P'}{P} = 128.$

(A) When the volume of an ideal gas is suddenly decreased,the process is adiabatic because there is no time for heat exchange with the surroundings $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$. Since the gas is compressed,work is done on the gas $(W < 0)$,which leads to an increase in internal energy $(\Delta U > 0)$.
Since the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$,the temperature of the gas increases. Thus,statement $III$ is false.
The average kinetic energy of gas atoms is proportional to the absolute temperature $(KE_{avg} = \frac{3}{2}kT)$,so the kinetic energy increases. Thus,statement $I$ is true.
Because the volume is decreased and the speed of the atoms increases,the atoms collide with the walls of the cylinder more frequently,leading to an increase in pressure. Thus,statement $II$ is true.
Therefore,statements $I$ and $II$ are true.

(B) The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_v \Delta T$.
For an ideal gas,$C_v = \frac{R}{\gamma - 1}$.
Given: $n = 1 \, mol$,$\gamma = 1.4$,$R = 8.3 \, J/mol/K$,$T_1 = 27\,^{\circ}C = 300 \, K$,$T_2 = 35\,^{\circ}C = 308 \, K$.
$\Delta T = T_2 - T_1 = 308 - 300 = 8 \, K$.
Substituting the values:
$\Delta U = 1 \times \frac{8.3}{1.4 - 1} \times 8$
$\Delta U = \frac{8.3}{0.4} \times 8$
$\Delta U = 8.3 \times 20 = 166 \, J$.

(B) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Using the ideal gas equation for $1$ mole,$PV = RT$,we can write $V = \frac{RT}{P}$.
Substituting this into the adiabatic equation: $P \left( \frac{RT}{P} \right)^{\gamma} = \text{constant}$.
This simplifies to $P^{1-\gamma} T^{\gamma} = \text{constant}$.
Therefore,$P_1^{1-\gamma} T_1^{\gamma} = P_2^{1-\gamma} T_2^{\gamma}$.
Rearranging for $T_2$: $\left( \frac{T_2}{T_1} \right)^{\gamma} = \left( \frac{P_1}{P_2} \right)^{1-\gamma} = \left( \frac{P_2}{P_1} \right)^{\gamma - 1}$.
Taking the $\gamma$-th root on both sides: $T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$.

(B) For an adiabatic process, the equation is $PV^{\gamma} = \text{constant}$.
Differentiating with respect to $V$, we get $\frac{dP}{dV} V^{\gamma} + P \gamma V^{\gamma-1} = 0$.
Thus, the slope of the $P-V$ curve is $\frac{dP}{dV} = -\gamma \frac{P}{V}$.
The magnitude of the slope is proportional to the adiabatic index $\gamma$ (i.e., $|\text{slope}| \propto \gamma$).
From the figure, the slope of curve $2$ is greater than the slope of curve $1$ (at any given $V$).
Therefore, $\gamma_2 > \gamma_1$.
For a monoatomic gas like $He$, $\gamma = 1.67$, and for a diatomic gas like $O_2$, $\gamma = 1.4$.
Since $\gamma_{mono} > \gamma_{dia}$, curve $2$ corresponds to a monoatomic gas $(He)$ and curve $1$ corresponds to a diatomic gas $(O_2)$.
Thus, plots $1$ and $2$ correspond to $O_2$ and $He$ respectively.

(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 $P_1$ and initial volume be $V_1$.
Let the final pressure be $P_2$ and final volume be $V_2 = \frac{V_1}{8}$.
Using the adiabatic relation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma}$.
Substituting the values: $P_2 = P_1 \left( \frac{V_1}{V_1/8} \right)^{5/3} = P_1 (8)^{5/3}$.
Since $8 = 2^3$, we have $P_2 = P_1 (2^3)^{5/3} = P_1 (2^5) = 32 P_1$.
Therefore, the pressure becomes $32$ times the initial pressure.

(C) Given: Work done on the gas,$W = -830 \ J$ (since work is done on the system).
Number of moles,$\mu = 2$.
For a diatomic ideal gas,the adiabatic exponent $\gamma = 1.4$.
Work done in an adiabatic process is given by the formula:
$W = \frac{\mu R (T_1 - T_2)}{\gamma - 1} = -\frac{\mu R \Delta T}{\gamma - 1}$
Substituting the values:
$-830 = -\frac{2 \times 8.3 \times \Delta T}{1.4 - 1}$
$830 = \frac{16.6 \times \Delta T}{0.4}$
$830 = 41.5 \times \Delta T$
$\Delta T = \frac{830}{41.5} = 20 \ K$
Thus,the change in temperature is $20 \ K$.

(B) For an adiabatic process,the relationship between pressure $P$ and density $\rho$ is given by $P \propto \rho^{\gamma}$.
Given that the initial pressure $P_1 = 1 \text{ atm}$ and the final pressure $P_2 = 128 \text{ atm}$.
The density changes from $\rho$ to $\rho' = 32\rho$.
Using the relation $\frac{P_2}{P_1} = \left(\frac{\rho_2}{\rho_1}\right)^{\gamma}$,we substitute the values:
$\frac{128}{1} = (32)^{\gamma}$.
We know that $128 = 2^7$ and $32 = 2^5$.
So,$2^7 = (2^5)^{\gamma} = 2^{5\gamma}$.
Equating the exponents: $7 = 5\gamma$.
Therefore,$\gamma = \frac{7}{5} = 1.4$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.
In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Therefore,$0 = \Delta U + W$,which implies $\Delta U = -W$ or $W = -\Delta U$.
This means the change in internal energy is equal to the negative of the work done by the gas (or equal to the work done on the gas). Thus,Statement $1$ is true.
In an adiabatic process,the temperature of a gas changes because the internal energy changes as work is done. Therefore,Statement $2$ is false.

(C) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Let the initial temperature be $T_1 = T$ and initial volume be $V_1 = V$.
The final volume is $V_2 = \frac{V}{4}$ and the adiabatic index is $\gamma = 1.5$.
Using the adiabatic relation:
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$
Substitute the given values:
$T (V)^{1.5-1} = T_2 (\frac{V}{4})^{1.5-1}$
$T (V)^{0.5} = T_2 (\frac{V}{4})^{0.5}$
Divide both sides by $V^{0.5}$:
$T = T_2 (\frac{1}{4})^{0.5}$
$T = T_2 (\frac{1}{2})$
Therefore,the final temperature $T_2 = 2T$.

(B) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} =$ constant.
Comparing this with the given equation $TV^x =$ constant,we get $x = \gamma - 1$.
For a rigid diatomic ideal gas,the degrees of freedom $f = 5$.
The adiabatic exponent $\gamma$ is given by $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5}$.
Substituting the value of $\gamma$ into the expression for $x$:
$x = \frac{7}{5} - 1 = \frac{2}{5}$.

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^\gamma = \text{constant}$.
From the ideal gas equation,$PV = \mu RT$,we have $V = \frac{\mu RT}{P}$.
Substituting this into the adiabatic equation: $P \left( \frac{\mu RT}{P} \right)^\gamma = \text{constant}$.
This simplifies to $P^{1-\gamma} T^\gamma = \text{constant}$,or $P \propto T^{\frac{\gamma}{\gamma-1}}$.
For a diatomic gas,the adiabatic exponent $\gamma = 7/5 = 1.4$.
Therefore,$C = \frac{\gamma}{\gamma-1} = \frac{7/5}{7/5 - 1} = \frac{7/5}{2/5} = 7/2$.

(C) When a tyre bursts suddenly,the process is adiabatic because the expansion happens very quickly,allowing no time for heat exchange with the surroundings.
For an adiabatic process,the relationship between temperature $T$ and pressure $P$ is given by $T^{\gamma} P^{1-\gamma} = \text{constant}$.
Thus,$T_1^{\gamma} P_1^{1-\gamma} = T_2^{\gamma} P_2^{1-\gamma}$,which can be rewritten as $T_2 = T_1 (P_1/P_2)^{(1-\gamma)/\gamma}$.
Given: $T_1 = 27\,^{\circ}C = 300\,K$,$P_1 = 2\,atm$,$P_2 = 1\,atm$ (atmospheric pressure),and $\gamma = 1.4$.
Substituting the values: $T_2 = 300 \times (2/1)^{(1-1.4)/1.4} = 300 \times (2)^{-0.4/1.4} = 300 \times (2)^{-2/7}$.
Calculating the value: $T_2 \approx 300 / 1.219 \approx 246\,K$.
Converting to Celsius: $T_2 = 246 - 273 = -27\,^{\circ}C$.

(B) For an adiabatic process, the equation is $PV^{\gamma} = \text{constant}$.
Differentiating with respect to $V$, we get $\frac{dP}{dV} = -\gamma \frac{P}{V}$.
The magnitude of the slope is $|\text{slope}| = \gamma \frac{P}{V}$.
Thus, the slope is directly proportional to the adiabatic index $\gamma$.
From the given graph, the magnitude of the slope of curve $(2)$ is greater than the magnitude of the slope of curve $(1)$.
Therefore, $\gamma_2 > \gamma_1$.
For a monoatomic gas (like $He$), $\gamma = \frac{5}{3} \approx 1.67$.
For a diatomic gas (like $O_2$), $\gamma = \frac{7}{5} = 1.4$.
Since $\gamma_{\text{mono}} > \gamma_{\text{dia}}$, curve $(2)$ corresponds to the monoatomic gas $(He)$ and curve $(1)$ corresponds to the diatomic gas $(O_2)$.
Thus, plots $(1)$ and $(2)$ correspond to $O_2$ and $He$ respectively.

(A) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
Taking the natural logarithm on both sides: $\ln P + \gamma \ln V = \text{constant}$.
Differentiating both sides: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.
This implies $\frac{dP}{P} = -\gamma \frac{dV}{V}$.
Given $\frac{dP}{P} = \frac{2}{3}\% $ and $\gamma = \frac{3}{2}$.
Substituting these values: $\frac{2}{3}\% = -\frac{3}{2} \frac{dV}{V}$.
Solving for the percentage change in volume: $\frac{dV}{V} = -\frac{2}{3} \times \frac{2}{3}\% = -\frac{4}{9}\%$.
The negative sign indicates a decrease in volume. Thus,the volume decreases by $\frac{4}{9}\% $.

(B) For an adiabatic process, the equation is $PV^{\gamma} = \text{constant}$.
Taking the derivative with respect to $V$, we get $P(\gamma V^{\gamma-1}) + V^{\gamma} \frac{dP}{dV} = 0$.
This simplifies to $\frac{dP}{dV} = -\gamma \frac{P}{V}$.
The magnitude of the slope of the $P-V$ curve is proportional to the adiabatic index $\gamma$, where $\gamma = \frac{C_p}{C_v}$.
For a monoatomic gas (like $He$), $\gamma = \frac{5}{3} \approx 1.67$.
For a diatomic gas (like $O_2$), $\gamma = \frac{7}{5} = 1.4$.
Since the slope of curve $(2)$ is greater than the slope of curve $(1)$, curve $(2)$ corresponds to the gas with a higher $\gamma$ value (monoatomic gas, $He$) and curve $(1)$ corresponds to the gas with a lower $\gamma$ value (diatomic gas, $O_2$).
Therefore, plots $(1)$ and $(2)$ correspond to $O_2$ and $He$ respectively.

(A) For an ideal gas,the number of moles $n$ at $STP$ is given by $n = \frac{V}{22.4} = \frac{5.6}{22.4} = 0.25 = \frac{1}{4} \, mol$.
Helium is a monoatomic gas,so the adiabatic index $\gamma = \frac{5}{3}$.
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: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = T_1 \left( \frac{5.6}{0.7} \right)^{\frac{5}{3}-1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4 T_1$.
The work done in an adiabatic process is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
Since it is compression,the magnitude of work done is $|W| = \frac{nR(T_2 - T_1)}{\gamma - 1}$.
Substituting the values: $|W| = \frac{(1/4) R (4 T_1 - T_1)}{(5/3) - 1} = \frac{(1/4) R (3 T_1)}{2/3} = \frac{3/4 R T_1}{2/3} = \frac{3}{4} \times \frac{3}{2} R T_1 = \frac{9}{8} R T_1$.

(A) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$.
Given that $P \propto T^3$,we can write $P = k T^3$,which implies $P T^{-3} = k$.
Comparing $P T^{-3} = k$ with the adiabatic relation $P T^{\frac{\gamma}{1-\gamma}} = k$,we get:
$\frac{\gamma}{1-\gamma} = -3$
$\gamma = -3(1-\gamma)$
$\gamma = -3 + 3\gamma$
$2\gamma = 3$
$\gamma = 3/2$.

(A) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Let the initial volume be $V_0$ and the initial temperature be $T_0 = 27 + 273 = 300 \, K$.
The final volume is $V = V_0 / 4$.
Given $\gamma = 1.4$, then $\gamma - 1 = 0.4$.
Using the adiabatic relation: $T_0 V_0^{\gamma-1} = T V^{\gamma-1}$.
$300 \times V_0^{0.4} = T \times (V_0 / 4)^{0.4}$.
$300 \times V_0^{0.4} = T \times V_0^{0.4} \times (1/4)^{0.4}$.
$300 = T \times (4^{-1})^{0.4}$.
$300 = T \times 4^{-0.4}$.
$T = 300 \times 4^{0.4} \, K$.

(B) For an adiabatic process,the relationship between volume $V$ and temperature $T$ is given by $TV^{\gamma-1} = \text{constant}$.
Given that $V \propto \frac{1}{T^3}$,we can write $V T^3 = \text{constant}$.
Raising this to the power of $1/3$,we get $V^{1/3} T = \text{constant}$.
Comparing this with the standard adiabatic equation $V^{\gamma-1} T = \text{constant}$,we have $\gamma - 1 = 1/3$.
Therefore,$\gamma = 1 + 1/3 = 4/3$.
Since $\gamma = \frac{C_p}{C_v}$,the ratio is $4/3 \approx 1.33$.

(D) For an adiabatic process, the relationship between pressure $P$ and volume $V$ is given by $P V^{\gamma} = \text{constant}$.
When the volume $V$ decreases during adiabatic compression, the pressure $P$ must increase to maintain the constant product.
Similarly, the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Since $\gamma > 1$ for all gases, $\gamma - 1 > 0$. As the volume $V$ decreases, the temperature $T$ must increase to keep the product constant.
Therefore, in adiabatic compression, both temperature and pressure increase.

(C) Since the air escapes suddenly from the puncture,the process is adiabatic.
In an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
This implies $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
As the air expands while escaping,the final volume $V_2$ is greater than the initial volume $V_1$ $(V_2 > V_1)$.
Therefore,the final temperature $T_2$ must be less than the initial temperature $T_1$ $(T_1 > T_2)$.
Thus,the air inside starts becoming cooler.

(D) For an adiabatic process,the relationship between pressure $P$ and volume $V$ is given by $P V^{\gamma} = \text{constant}$.
From the ideal gas equation,$PV = nRT$,we can express volume as $V = \frac{nRT}{P}$.
Substituting this expression for $V$ into the adiabatic equation:
$P \left( \frac{nRT}{P} \right)^{\gamma} = \text{constant}$.
Since $n$ and $R$ are constants,we have $P \cdot \frac{T^{\gamma}}{P^{\gamma}} = \text{constant}$.
This simplifies to $P^{1 - \gamma} T^{\gamma} = \text{constant}$.

(C) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, so $\Delta U = -W$.
Given that the work done by the gas is $W = 6 \, R \, \text{J}$, we have $\Delta U = -6 \, R$.
The change in internal energy for an ideal gas is given by $\Delta U = n C_V \Delta T$.
For a gas with adiabatic index $\gamma = \frac{5}{3}$, the molar heat capacity at constant volume is $C_V = \frac{R}{\gamma - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2} R$.
Substituting the values, we get $-6 \, R = 1 \times \left( \frac{3}{2} R \right) \times (T_{final} - T)$.
Dividing both sides by $R$, we get $-6 = \frac{3}{2} (T_{final} - T)$.
Solving for the temperature change: $T_{final} - T = -6 \times \frac{2}{3} = -4$.
Therefore, $T_{final} = (T - 4) \, K$.

(D) In an adiabatic process,the system is thermally insulated from the surroundings,meaning $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$.
For adiabatic compression,work is done on the system,so $dW < 0$.
This implies $dU = -dW > 0$,which means the internal energy $U$ increases.
Since internal energy of an ideal gas is a function of temperature $(U \propto T)$,the temperature of the system also increases.
Therefore,the $Assertion$ is incorrect.
Adiabatic processes are typically fast processes to ensure no heat exchange occurs with the surroundings.
Therefore,the $Reason$ is also incorrect.
Thus,both $Assertion$ and $Reason$ are incorrect.

(A) When a bottle of cold carbonated drink is opened,the high-pressure gas inside the bottle expands rapidly into the atmosphere.
This rapid expansion is an adiabatic process because it happens so quickly that there is no time for heat exchange with the surroundings.
According to the first law of thermodynamics,for an adiabatic expansion $(dQ = 0)$,the work done by the gas $(dW > 0)$ results in a decrease in internal energy $(dU < 0)$,which leads to a significant drop in temperature.
Due to this sudden cooling,the water vapour present in the air near the opening condenses into tiny liquid droplets,forming a visible fog.

(A) When air leaks out of a balloon quickly,the process happens so fast that there is no time for heat exchange with the surroundings,making it an adiabatic process.
During this process,the air expands against the external atmospheric pressure.
Since the air does work at the expense of its internal energy,its temperature decreases,causing the air to cool down.

(C) For an adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Since $\gamma > 1$ for all gases, $T \propto V^{-(\gamma-1)}$.
In an adiabatic expansion, the volume $V$ increases, which implies that the temperature $T$ must decrease. Thus, the Assertion is correct.
However, the Reason states that volume is inversely proportional to temperature $(V \propto 1/T)$, which is incorrect. The correct relation is $T \propto V^{-(\gamma-1)}$.
Therefore, the Assertion is correct but the Reason is incorrect.

(B) In an adiabatic process,the system is thermally insulated from its surroundings.
Therefore,there is no exchange of heat between the system and the surroundings.
This means that the heat absorbed or released by the system is zero.
Mathematically,this is represented as $\Delta Q = 0$.

(A) The mean collision time $\tau$ is defined as the ratio of the mean free path $\lambda$ to the root mean square speed $v_{RMS}$.
$\tau = \frac{\lambda}{v_{RMS}}$
We know that the mean free path $\lambda = \frac{V}{\sqrt{2} \pi d^2 N}$,so $\lambda \propto V$.
Also,$v_{RMS} = \sqrt{\frac{3RT}{M}} \propto \sqrt{T}$.
From the ideal gas law $PV = nRT$,we have $T \propto PV$,so $v_{RMS} \propto \sqrt{PV}$.
Substituting these into the expression for $\tau$:
$\tau \propto \frac{V}{\sqrt{PV}} = \sqrt{\frac{V^2}{PV}} = \sqrt{\frac{V}{P}}$.
For an adiabatic process,$PV^{\gamma} = \text{constant}$,which implies $P \propto V^{-\gamma}$.
Substituting this into the proportionality for $\tau$:
$\tau \propto \sqrt{\frac{V}{V^{-\gamma}}} = \sqrt{V^{1+\gamma}} = V^{\frac{1+\gamma}{2}}$.
Given the volume changes from $V_1$ to $V_2 = 2V_1$,the ratio is:
$\frac{\tau_2}{\tau_1} = \left(\frac{V_2}{V_1}\right)^{\frac{1+\gamma}{2}} = (2)^{\frac{1+\gamma}{2}}$.
Since the question asks for the ratio in terms of $\frac{1}{2}$,we have $\frac{\tau_2}{\tau_1} = \left(\frac{1}{2}\right)^{-\frac{\gamma+1}{2}}$.
Note: Based on the provided options,the intended answer is $\left(\frac{1}{2}\right)^{-\frac{\gamma+1}{2}}$,but since the options are formatted as powers of $1/2$,the correct mathematical derivation leads to the inverse of option $A$.

(A) For an adiabatic process,the work done is given by $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
At $STP$,$P_1 = 1.013 \times 10^5 \; Pa$ and $V_1 = 1 \; L = 10^{-3} \; m^3$.
Using the adiabatic relation $P_1 V_1^\gamma = P_2 V_2^\gamma$,we find $P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = P_1 \left(\frac{1}{3}\right)^{1.4}$.
Substituting $P_2$ into the work formula: $W = \frac{P_1 V_1 - P_1 V_1 (1/3)^{1.4} \times 3}{\gamma - 1} = \frac{P_1 V_1 [1 - 3 \times (1/3)^{1.4}]}{0.4}$.
Given $3^{1.4} = 4.6555$,then $(1/3)^{1.4} = 1/4.6555 \approx 0.2148$.
$W = \frac{1.013 \times 10^5 \times 10^{-3} \times [1 - 3 \times 0.2148]}{0.4} = \frac{101.3 \times [1 - 0.6444]}{0.4} = \frac{101.3 \times 0.3556}{0.4} \approx 90.04 \; J$.
Rounding to the nearest provided option,the work done is approximately $90.5 \; J$.

(2.639) The cylinder is completely insulated from its surroundings. As a result,no heat is exchanged between the system (cylinder) and its surroundings. Thus,the process is adiabatic.
Initial pressure inside the cylinder $= P_1$
Final pressure inside the cylinder $= P_2$
Initial volume inside the cylinder $= V_1$
Final volume inside the cylinder $= V_2 = V_1 / 2$
Ratio of specific heats for hydrogen (diatomic gas),$\gamma = 1.4$
For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
Substituting the values:
$P_1 V_1^{1.4} = P_2 (V_1 / 2)^{1.4}$
$P_2 / P_1 = (V_1 / (V_1 / 2))^{1.4}$
$P_2 / P_1 = 2^{1.4}$
$P_2 / P_1 \approx 2.639$
Hence,the pressure of the gas increases by a factor of $2.639$.

(A) Yes,it is possible to increase the temperature of a gas without adding heat to it. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. If we perform adiabatic compression on the gas,then $\Delta Q = 0$. In this case,$\Delta U = -\Delta W$. Since work is done on the gas $(\Delta W < 0)$,the internal energy $\Delta U$ increases. Because the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$,an increase in internal energy leads to an increase in the temperature of the gas.

(A) In an adiabatic process,the work done $W$ by the gas is given by the formula $W = \frac{P_{i}V_{i} - P_{f}V_{f}}{\gamma - 1}$.
Since the volume $V$ of the tube remains constant throughout the process,we have $V_{i} = V_{f} = V$.
The initial pressure is $P_{1}$ and the final pressure is $P_{2}$.
Substituting these values into the work formula,we get $W = \frac{P_{1}V - P_{2}V}{\gamma - 1}$.
However,the work done $ON$ the gas (by the pump) is the negative of the work done $BY$ the gas.
Therefore,the work done to increase the pressure is $W_{ext} = -W = -\left(\frac{P_{1}V - P_{2}V}{\gamma - 1}\right) = \frac{P_{2}V - P_{1}V}{\gamma - 1}$.

(N/A) An adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings,i.e.,$\Delta Q = 0$.
From the first law of thermodynamics:
$\Delta Q = \Delta U + W$
Since $\Delta Q = 0$,we have $W = -\Delta U$.
For an ideal gas undergoing an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = K$ (constant),where $\gamma = \frac{C_P}{C_V}$ is the adiabatic index.
The work done $W$ during the expansion from state $(P_1, V_1)$ to $(P_2, V_2)$ is:
$W = \int_{V_1}^{V_2} P \, dV$
Since $P = K V^{-\gamma}$,we substitute this into the integral:
$W = \int_{V_1}^{V_2} K V^{-\gamma} \, dV = K \left[ \frac{V^{-\gamma+1}}{-\gamma+1} \right]_{V_1}^{V_2}$
$W = \frac{K}{1-\gamma} (V_2^{1-\gamma} - V_1^{1-\gamma})$
Since $K = P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$,we get:
$W = \frac{1}{1-\gamma} (P_2 V_2^{\gamma} V_2^{1-\gamma} - P_1 V_1^{\gamma} V_1^{1-\gamma})$
$W = \frac{P_2 V_2 - P_1 V_1}{1-\gamma} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$
Using the ideal gas law $PV = nRT$,the expression can also be written as:
$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$

(B) In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$,where $dU$ is the change in internal energy and $dW$ is the work done.
Since $dQ = 0$,we have $0 = dU + dW$,which implies $dU = -dW$.
When work is done by the gas,$dW > 0$.
Therefore,$dU = -dW < 0$,meaning the internal energy of the gas decreases.
Since the internal energy of an ideal gas is directly proportional to its temperature $(U \propto T)$,a decrease in internal energy leads to a decrease in the temperature of the gas.

(A) An adiabatic process is defined as a process in which there is no exchange of heat between the system and its surroundings $(Q = 0)$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
If $Q = 0$,then $\Delta U = -W$.
This means that a change in internal energy $(\Delta U)$ can occur during an adiabatic process due to work done by or on the system.
Therefore,a change in internal energy is a characteristic feature of an adiabatic process,provided the work done is non-zero.

(A) Yes,the temperature of an isolated system of gas can be changed.
An isolated system is one that does not exchange matter or heat with its surroundings.
However,work can still be done on or by the system.
According to the first law of thermodynamics,$\Delta U = Q - W$.
For an isolated system,$Q = 0$,so $\Delta U = -W$.
If the gas is compressed adiabatically $(W < 0)$,the internal energy $\Delta U$ increases,which leads to an increase in temperature.
Conversely,if the gas expands adiabatically $(W > 0)$,the internal energy decreases,leading to a decrease in temperature.

(A) When the valve of a bicycle tube is opened,the compressed air inside the tube expands rapidly to the outside atmosphere. This process occurs so quickly that there is no time for heat exchange with the surroundings,making it an adiabatic process. According to the first law of thermodynamics,$dQ = dU + dW$. Since the process is adiabatic,$dQ = 0$,which implies $dU = -dW$. As the air does work $(dW > 0)$ in expanding against the external atmospheric pressure,its internal energy $(dU)$ decreases. Since the internal energy of an ideal gas is directly proportional to its temperature,the decrease in internal energy leads to a drop in temperature,making the escaping air feel cold.

(N/A) As air rises,the atmospheric pressure decreases. Due to this decrease in pressure,the air parcel expands. Since the expansion occurs rapidly,it can be considered an adiabatic process where the air does work on the surroundings at the expense of its internal energy. According to the first law of thermodynamics,$dQ = dU + dW$. For an adiabatic process,$dQ = 0$,so $dU = -dW$. Since the air does positive work $(dW > 0)$,its internal energy decreases $(dU < 0)$,which leads to a decrease in temperature.

(C) An adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings.
Therefore,the total heat content $(Q)$ of the system remains constant during the process,meaning $\Delta Q = 0$.

(N/A) When water passes through the small holes of a shower head,it undergoes a rapid expansion. According to the principle of thermodynamics,this rapid expansion is approximately an adiabatic process. During an adiabatic expansion,the internal energy of the gas or fluid decreases,leading to a drop in temperature. In summer,the ambient temperature is high,so the cooling effect of the water makes it feel pleasant. However,in winter,the ambient temperature is already low,and the additional cooling caused by the adiabatic expansion makes the water feel uncomfortably cold.