Adiabatic Process Questions in English

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Given that heat is not exchanged,$\Delta Q = 0$.
Therefore,$0 = \Delta U + \Delta W$,which implies $\Delta W = -\Delta U$.
If the internal energy $\Delta U$ increases,the system does negative work (work is done on the system).
For an ideal gas,internal energy $U$ is a function of temperature $T$ only $(U \propto T)$.
Since internal energy increases,the temperature of the body must increase.

(C) For an adiabatic process,the work done $W$ by a gas is given by the formula:
$W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$
where $\mu$ is the number of moles,$R$ is the universal gas constant,$T_1$ and $T_2$ are the initial and final temperatures,and $\gamma$ is the adiabatic index.
Since $\mu$,$R$,and $\gamma$ are constants for a given gas,the work done depends directly on the change in temperature $(T_1 - T_2)$.

(B) In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
According to the First Law of Thermodynamics $(FLOT)$,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta Q = 0$,we have $0 = \Delta U + \Delta W$,which implies $\Delta U = -\Delta W$.
In an expansion process,the system does work on the surroundings,so the work done by the system $\Delta W$ is positive.
Therefore,$\Delta U = -(\text{positive value})$,which means $\Delta U$ is negative.

(D) sudden bursting of a tyre is an adiabatic process 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 $T^\gamma P^{1-\gamma} = \text{constant}$.
This can be written as $T_1^\gamma P_1^{1-\gamma} = T_2^\gamma P_2^{1-\gamma}$.
Rearranging for $T_2$,we get $\frac{T_2}{T_1} = \left( \frac{P_1}{P_2} \right)^{\frac{1-\gamma}{\gamma}}$.
Given: $T_1 = 300 \ K$,$P_1 = 4P_{atm}$,$P_2 = P_{atm}$,and $\gamma = 1.4$.
Substituting the values: $\frac{T_2}{300} = \left( \frac{4P_{atm}}{P_{atm}} \right)^{\frac{1-1.4}{1.4}}$.
$\frac{T_2}{300} = (4)^{\frac{-0.4}{1.4}}$.
Therefore,$T_2 = 300 \ (4)^{-0.4/1.4}$.

(C) For a sudden compression,the process is adiabatic.
For an adiabatic process,the relation between pressure and volume is given by $P V^{\gamma} = \text{constant}$.
Thus,$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$,which implies $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}$.
Given: Initial pressure $P_1 = 1 \text{ atm}$ (at $NTP$),final volume $V_2 = \frac{V_1}{4}$,and $\gamma = \frac{3}{2}$.
Substituting these values: $\frac{P_2}{1} = \left( \frac{V_1}{V_1/4} \right)^{3/2} = (4)^{3/2}$.
Calculating the value: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,the final pressure $P_2 = 8 \text{ atm}$.

(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 $V_2 = V_1/8$.
Using the adiabatic relation: $P_1 V_1^\gamma = P_2 V_2^\gamma$.
$P_2 = P_1 (V_1 / V_2)^\gamma$.
Substituting the values: $P_2 = P_1 (V_1 / (V_1/8))^{5/3}$.
$P_2 = P_1 (8)^{5/3}$.
$P_2 = P_1 (2^3)^{5/3} = P_1 (2^5) = 32 P_1$.
Therefore,the pressure becomes $32$ times its initial pressure.

(C) For an adiabatic process,the relation between pressure $P$ and density $d$ is given by $P \propto d^\gamma$.
Therefore,the ratio of pressures is $\frac{P'}{P} = \left( \frac{d'}{d} \right)^\gamma$.
Given that $\frac{d'}{d} = 32$ and $\gamma = 7/5$,we substitute these values into the equation:
$\frac{P'}{P} = (32)^{7/5}$.
Since $32 = 2^5$,we have $\frac{P'}{P} = (2^5)^{7/5} = 2^{(5 \times 7/5)} = 2^7$.
Calculating $2^7$,we get $2^7 = 128$.
Thus,the ratio $\frac{P'}{P}$ is $128$.

(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: Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.
Final volume $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.
Adiabatic index $\gamma = \frac{5}{3}$,so $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Using the formula $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$:
$T_2 = 300 \times \left( \frac{27}{8} \right)^{2/3}$.
$T_2 = 300 \times \left( \left( \frac{27}{8} \right)^{1/3} \right)^2 = 300 \times \left( \frac{3}{2} \right)^2$.
$T_2 = 300 \times \frac{9}{4} = 75 \times 9 = 675 \ K$.
The rise in temperature is $\Delta T = T_2 - T_1 = 675 - 300 = 375 \ K$.

(D) For an adiabatic process,the relationship between pressure $(P)$ and volume $(V)$ is given by $PV^\gamma = \text{constant}$.
From the ideal gas equation,$PV = RT$,we can write $V = \frac{RT}{P}$.
Substituting this into the adiabatic equation:
$P \left( \frac{RT}{P} \right)^\gamma = \text{constant}$
$P \cdot \frac{T^\gamma}{P^\gamma} = \text{constant}$
$P^{1-\gamma} T^\gamma = \text{constant}$.
Therefore,the correct option is $D$.

(B) For an adiabatic process,the work done $W$ by an ideal gas is given by the first law of thermodynamics: $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$,so $W = -\Delta U$.
For an ideal gas,the change in internal energy is $\Delta U = nC_v \Delta T$. For $1 \text{ mole}$ of gas,$n = 1$ and $C_v = \frac{R}{\gamma - 1}$.
Thus,$W = -[C_v(T_1 - T)] = -[\frac{R}{\gamma - 1}(T_1 - T)]$.
Simplifying this,we get $W = \frac{R}{\gamma - 1}(T - T_1)$.

(D) For an adiabatic process,the heat exchange $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$,where $dU$ is the change in internal energy and $dW$ is the work done by the gas.
Given that the internal energy decreases by $2 \ J$,we have $dU = -2 \ J$.
Substituting these values into the first law equation: $0 = -2 \ J + dW$.
This gives $dW = 2 \ J$,which is the work done $BY$ the gas.
The question asks for the work done $ON$ the gas.
Work done $ON$ the gas $= -dW = -2 \ J$.

(B) For an isothermal process, the equation is $PV = \text{constant}$. Differentiating with respect to $V$, we get $P + V(dP/dV) = 0$, so the slope of the isothermal curve is $(dP/dV)_{\text{iso}} = -P/V$.
For an adiabatic process, the equation is $PV^{\gamma} = \text{constant}$. Differentiating with respect to $V$, we get $P(\gamma V^{\gamma-1}) + V^{\gamma}(dP/dV) = 0$, which simplifies to $(dP/dV)_{\text{adia}} = -\gamma P/V$.
The ratio of the slopes is $\frac{(dP/dV)_{\text{adia}}}{(dP/dV)_{\text{iso}}} = \frac{-\gamma P/V}{-P/V} = \gamma$.

(C) For an adiabatic process,the heat exchange $\Delta Q = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta Q = 0$,we have $0 = \Delta U + \Delta W$,which implies $\Delta W = -\Delta U$.
When a gas expands,it does work on the surroundings,so $\Delta W > 0$.
Consequently,$\Delta U$ must be negative,meaning the internal energy of the gas decreases as it is used to perform work.

(A) For an adiabatic process,the work done $W$ is given by the formula: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
Given: $n = 1 \, \text{mole}$,$R = 8.31 \, J/(mol \cdot K)$,$\gamma = 1.4$,$T_1 = 27^{\circ}C = 300 \, K$,and $T_2 = 127^{\circ}C = 400 \, K$.
Substituting the values: $W = \frac{1 \times 8.31 \times (300 - 400)}{1.4 - 1}$.
$W = \frac{8.31 \times (-100)}{0.4} = \frac{-831}{0.4} = -2077.5 \, J$.
The negative sign indicates that work is done on the gas during compression. The magnitude of the work done is $2077.5 \, J$.

(C) When compressed air escapes suddenly from a puncture,the process is adiabatic because the expansion happens very quickly,allowing no time for heat exchange with the surroundings $(dQ = 0)$. According to the first law of thermodynamics,$dU = dQ - dW$. Since the gas does work $(dW > 0)$ in expanding against the external atmosphere,the internal energy $(dU)$ decreases. As the internal energy of an ideal gas is directly proportional to its temperature,the temperature of the air inside the tube decreases.

(D) For an adiabatic process,the relationship between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$,where $\gamma$ is the adiabatic index.
Differentiating both sides with respect to $V$,we get: $P(\gamma V^{\gamma-1}) + V^{\gamma} \frac{dP}{dV} = 0$.
Rearranging the terms: $V^{\gamma} \frac{dP}{dV} = -\gamma P V^{\gamma-1}$.
Dividing by $V^{\gamma-1}$,we get: $V \frac{dP}{dV} = -\gamma P$.
The bulk modulus $B$ is defined as $B = -V \frac{dP}{dV}$.
Substituting the expression,we get $B = -(-\gamma P) = \gamma P$.
Therefore,the adiabatic bulk modulus is $\gamma P$.

(C) In an adiabatic process,the system is thermally insulated from its surroundings.
By definition,there is no exchange of heat between the system and the surroundings,which means $dQ = 0$.
Therefore,the heat content of the system remains constant during the process.

(B) For an adiabatic process,the relationship between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: Initial pressure $P_1 = P_0$,initial volume $V_1 = V$,final volume $V_2 = \frac{V}{8}$,and adiabatic index $\gamma = \frac{4}{3}$.
Substituting these values into the equation:
$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma$
$P_2 = P_0 \left( \frac{V}{V/8} \right)^{4/3}$
$P_2 = P_0 (8)^{4/3}$
$P_2 = P_0 (2^3)^{4/3}$
$P_2 = P_0 (2)^4$
$P_2 = 16P_0$.

(C) For an adiabatic process,the relationship between pressure $P$ and volume $V$ is given by $P V^\gamma = \text{constant}$.
Using the ideal gas equation $PV = RT$,we can express pressure as $P = \frac{RT}{V}$.
Substituting this into the adiabatic equation: $\left( \frac{RT}{V} \right) V^\gamma = \text{constant}$.
Since $R$ is a gas constant,we get $T V^{\gamma - 1} = \text{constant}$.

(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = \text{constant}$.
Therefore,we can write: $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.
Rearranging for the final temperature $T_2$: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}$.
Given: $T_1 = 300 \ K$,$V_2 = 2 V_1$,and $\gamma = 1.40$.
Substituting the values: $T_2 = 300 \left( \frac{V_1}{2 V_1} \right)^{1.40 - 1}$.
$T_2 = 300 \left( \frac{1}{2} \right)^{0.4}$.
$T_2 = 300 \times (0.5)^{0.4} \approx 300 \times 0.7578 = 227.36 \ K$.

(D) For an adiabatic process,the heat exchange $\Delta Q$ is equal to $0$.
According to the First Law of Thermodynamics: $\Delta Q = \Delta U + \Delta W$.
Since $\Delta Q = 0$,we have $\Delta W = -\Delta U$.
Given that the internal energy decreased by $100 \, J$,we have $\Delta U = -100 \, J$.
Substituting this value into the equation: $\Delta W = -(-100 \, J) = +100 \, J$.
Therefore,the work done by the gas is $100 \, J$.

(A) For an ideal gas,the change in internal energy $\Delta U$ is given by the formula $\Delta U = nC_V \Delta T$.
For $1$ mole of gas,$n = 1$.
The molar heat capacity at constant volume is $C_V = \frac{R}{\gamma - 1}$.
Therefore,$\Delta U = 1 \times \left( \frac{R}{\gamma - 1} \right) \times (T_2 - T_1)$.
Thus,the change in internal energy is $\Delta U = \frac{R}{\gamma - 1}(T_2 - T_1)$.

(D) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.
Given: $T_1 = 27^oC = 27 + 273 = 300 \ K$,$V_1 = 8 \ L$,$V_2 = 1 \ L$,and $\gamma = 5/3$.
Substituting the values: $300 \times (8)^{(5/3 - 1)} = T_2 \times (1)^{(5/3 - 1)}$.
$300 \times (8)^{2/3} = T_2$.
$300 \times (2^3)^{2/3} = T_2$.
$300 \times 2^2 = T_2$.
$300 \times 4 = 1200 \ K$.
To convert to Celsius: $T_2(^oC) = 1200 - 273 = 927^oC$.

(D) The process of a tyre bursting is extremely rapid. Due to the very short time interval,there is no sufficient time for the exchange of heat between the gas inside the tyre and the surroundings $(dQ = 0)$. According to the first law of thermodynamics,a process where no heat is exchanged with the surroundings is defined as an adiabatic process. Therefore,the correct option is $(d)$.

(A) The change in internal energy $\Delta U$ for an ideal gas is given by the formula $\Delta U = n C_V \Delta T$,where $n$ is the number of moles,$C_V$ is the molar heat capacity at constant volume,and $\Delta T = T_f - T_i$.
Given $n = 1$ mole of helium.
Therefore,$\Delta U = 1 \times C_V \times (T_f - T_i) = -C_V(T_i - T_f)$.
The decrease in internal energy is the magnitude of the change in internal energy,which is $|\Delta U| = C_V(T_i - T_f)$.

(C) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $T_1 = 273 \ K$,$V_2 = V_1/2$,and $\gamma = 1.41$.
$T_2 = T_1 (V_1/V_2)^{\gamma-1} = 273 \times (2)^{0.41} \approx 273 \times 1.328 = 362.54 \ K$.
The work done by the gas is $W = \frac{R(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{8.314 \times (273 - 362.54)}{1.41 - 1} = \frac{8.314 \times (-89.54)}{0.41} \approx -1815.6 \ J$.
The work done on the gas is $|W| \approx 1815 \ J$.

(D) In an adiabatic process,there is no exchange of heat between the system and the surroundings,which means $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$. Substituting $dQ = 0$,we get $0 = dU + dW$,or $dU = -dW$.
Since $dQ = 0$,the total heat $Q$ remains constant.
For a reversible adiabatic process,the entropy change $dS = dQ/T = 0$,meaning entropy remains constant. Therefore,the statement that entropy is not constant is incorrect.

(C) 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: Initial temperature $T_1 = 18 + 273 = 291 \ K$,initial volume $V_1 = V$,final volume $V_2 = V/8$,and for a diatomic gas,the adiabatic index $\gamma = 1.4$.
Substituting the values: $T_2 = T_1 (V_1 / V_2)^{\gamma - 1}$.
$T_2 = 291 \times (V / (V/8))^{1.4 - 1} = 291 \times (8)^{0.4}$.
Since $8^{0.4} = (2^3)^{0.4} = 2^{1.2} \approx 2.297$.
$T_2 = 291 \times 2.297 \approx 668.4 \ K$.
Converting to Celsius: $668.4 - 273 = 395.4^oC$. Note: Based on the provided options,$668 \ K$ is the correct value.

(A) The formula for specific heat capacity $c$ is given by $c = \frac{\Delta Q}{m \Delta T}$.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $\Delta Q = 0$.
Substituting this into the formula,we get $c = \frac{0}{m \Delta T} = 0$.
Therefore,the specific heat of a gas during an adiabatic process is zero.

(D) 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,meaning $dQ = 0$.
Thus,the correct option is $D$.

(B) For an adiabatic process,the work done by the gas is given by $W = \frac{\mu R(T_1 - T_2)}{\gamma - 1}$.
Using the adiabatic relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,we have $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.
Substituting this into the work formula: $W = \frac{\mu R T_1}{\gamma - 1} \left[ 1 - \left(\frac{V_1}{V_2}\right)^{\gamma-1} \right]$.
Given: $\mu = 2 \text{ mol}$,$T_1 = 27 + 273 = 300 \text{ K}$,$V_1 = V$,$V_2 = 2V$,$\gamma = 5/3$,$R = 8.31 \text{ J/mol K}$.
$W = \frac{2 \times 8.31 \times 300}{(5/3 - 1)} \left[ 1 - \left(\frac{V}{2V}\right)^{5/3 - 1} \right]$.
$W = \frac{4986}{2/3} \left[ 1 - (1/2)^{2/3} \right]$.
$W = 7479 \times [1 - 0.62996] = 7479 \times 0.37004 \approx 2767.23 \text{ J}$.

(D) For an adiabatic process,the relationship between temperature $T$ and pressure $P$ is given by $T^\gamma P^{1-\gamma} = \text{constant}$.
This can be rewritten as $T \propto P^{\frac{\gamma - 1}{\gamma}}$.
Given: Initial temperature $T_1 = 27^\circ C = 300K$,final pressure $P_2 = \frac{1}{8}P_1$,and $\gamma = 5/3$.
Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$.
Substituting the values: $\frac{T_2}{300} = \left( \frac{1}{8} \right)^{\frac{5/3 - 1}{5/3}} = \left( \frac{1}{8} \right)^{\frac{2/3}{5/3}} = \left( \frac{1}{8} \right)^{2/5}$.
$T_2 = 300 \times (8^{-1})^{2/5} = 300 \times (2^3)^{-2/5} = 300 \times 2^{-6/5} \approx 300 \times 0.435 = 131K$.
Converting to Celsius: $T_2 = 131 - 273 = -142^\circ C$.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ and the work done $\Delta W$ are related to the heat exchanged $\Delta Q$ as: $\Delta Q = \Delta U + \Delta W$.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,meaning $\Delta Q = 0$.
Substituting $\Delta Q = 0$ into the first law equation,we get: $0 = \Delta U + \Delta W$,or $\Delta U + \Delta W = 0$.
Therefore,the given equation is valid for an adiabatic process.

(D) For an adiabatic process,the relationship between temperature and pressure is given by $\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$.
Given: $P_1 = 1 \text{ atm}$,$P_2 = 8 \text{ atm}$,$T_1 = 27^{\circ}C = 300 \text{ K}$,and $\gamma = 3/2$.
Substituting the values: $\frac{T_2}{300} = (8)^{\frac{3/2 - 1}{3/2}} = (8)^{\frac{1/2}{3/2}} = (8)^{1/3}$.
Since $8^{1/3} = 2$,we have $\frac{T_2}{300} = 2$.
Therefore,$T_2 = 600 \text{ K}$.
Converting back to Celsius: $T_2(^{\circ}C) = 600 - 273 = 327^{\circ}C$.

(C) When the tube suddenly bursts,the process is adiabatic.
For an adiabatic process,the relationship between temperature and pressure is given by: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}$.
Here,the initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 K$.
The initial pressure $P_1 = 8 \text{ atm}$.
After the tube bursts,the final pressure $P_2 = 1 \text{ atm}$ (atmospheric pressure).
The adiabatic index $\gamma = 1.5$.
Substituting the values: $\frac{T_2}{300} = \left(\frac{1}{8}\right)^{\frac{1.5 - 1}{1.5}} = \left(\frac{1}{8}\right)^{\frac{0.5}{1.5}} = \left(\frac{1}{8}\right)^{\frac{1}{3}}$.
Since $8 = 2^3$,we have $\left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$.
Therefore,$T_2 = 300 \times \frac{1}{2} = 150 K$.

(C) For an adiabatic process, the relation between pressure and volume is given by $PV^\gamma = \text{constant}$.
Therefore, $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: $P_1 = P$, $V_2 = \frac{V_1}{8}$, and $\gamma = 2.5 = \frac{5}{2}$.
Substituting these values into the equation:
$P \times V_1^{5/2} = P' \times \left( \frac{V_1}{8} \right)^{5/2}$.
$P' = P \times \left( \frac{V_1}{V_1/8} \right)^{5/2}$.
$P' = P \times (8)^{5/2}$.
Since $8 = 2^3$, we have $P' = P \times (2^3)^{5/2} = P \times 2^{15/2}$.

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

(B) In an adiabatic process,there is no exchange of heat with 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 by the gas.
Since $dQ = 0$,we have $dU = -dW$.
In an expansion,the gas does work on the surroundings,so $dW > 0$.
This implies $dU < 0$,which means 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 temperature.
Therefore,in an adiabatic expansion,the temperature of the gas falls.

(D) For an adiabatic process,the work done $W$ by $n$ moles of an ideal gas is given by the formula:
$W = \frac{nR(T_i - T_f)}{\gamma - 1}$
Given:
$n = 1 \text{ mole}$
$T_i = T \ K$
$W = 6R \ J$
$\gamma = 5/3$
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 $R$ and solving for $T_f$:
$6 = 1.5(T - T_f)$
$T - T_f = 6 / 1.5 = 4$
$T_f = T - 4$
Therefore,the final temperature of the gas is $(T - 4) \ K$.

(A) For an adiabatic process,the relation between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.
Given: Initial volume $V_1 = V$,final volume $V_2 = V/4$,initial temperature $T_1 = 273 \, K$,and $\gamma = 1.5$.
Using the adiabatic relation: $T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$.
Substituting the values: $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} = T_1 \left( \frac{V}{V/4} \right)^{1.5 - 1} = T_1 (4)^{0.5} = T_1 \times 2$.
Thus,$T_2 = 2 \times 273 = 546 \, K$.
The increase in temperature is $\Delta T = T_2 - T_1 = 546 - 273 = 273 \, K$.

(B) Since the process is sudden and the vessel is insulated,it is an adiabatic process. For an adiabatic process,the relation between pressure and volume is $PV^\gamma = \text{constant}$.
Thus,$P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given: $P_1 = 10^5 \ N/m^2$,$\gamma = 1.3$,and $V_2 = \frac{V_1}{2}$.
Substituting these values into the equation:
$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma$
$P_2 = 10^5 \times \left( \frac{V_1}{V_1/2} \right)^{1.3}$
$P_2 = 10^5 \times (2)^{1.3} \ N/m^2$.
Therefore,the final pressure is $2^{1.3} \times 10^5 \ N/m^2$.

(B) For an adiabatic process,the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$. Since $\Delta Q = 0$ for an adiabatic process,we have $\Delta U = -\Delta W$.
In the case of compression,work is done on the gas,so $\Delta W$ is negative $(\Delta W < 0)$.
Substituting this into the equation,$\Delta U = -(\text{negative value})$,which results in $\Delta U > 0$.
Therefore,the internal energy of the gas increases during adiabatic compression.

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

(B) For a sudden compression,the process is adiabatic. The adiabatic equation is $P V^{\gamma} = \text{constant}$.
Let the initial pressure be $P_1 = P$ and initial volume be $V_1 = V$.
The final volume is $V_2 = V/4$.
Using the relation $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}$,we get:
$\frac{P_2}{P} = \left( \frac{V}{V/4} \right)^{\gamma} = 4^{\gamma}$.
Therefore,$P_2 = 4^{\gamma} P$.
Since the adiabatic index $\gamma$ for any gas is greater than $1$ (e.g.,$1.4$ for diatomic,$1.67$ for monatomic),$4^{\gamma}$ will be greater than $4$.
Thus,$P_2 > 4P$,which means the final pressure is much more than $P$.

(D) For an adiabatic process,the relation between pressure and volume is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$.
Given that $\gamma = 1.5 = \frac{3}{2}$ and the final volume $V_2 = \frac{V_1}{4}$,we have $\frac{V_1}{V_2} = 4$.
The ratio of final pressure to initial pressure is $\frac{P_2}{P_1} = \left( \frac{V_1}{V_2} \right)^\gamma$.
Substituting the values,$\frac{P_2}{P_1} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Thus,the ratio of the final pressure to the initial pressure is $8:1$.

(B) The change in internal energy $\Delta U$ of 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: $n = 1 \, mol$,$\gamma = 1.4$,$R = 8.3 \, J/mol \cdot K$.
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.
Final temperature $T_2 = 35^{\circ}C = 35 + 273 = 308 \, K$.
Change in temperature $\Delta T = T_2 - T_1 = 308 - 300 = 8 \, K$.
Substituting the values:
$\Delta U = 1 \times \frac{8.3}{1.4 - 1} \times 8$
$\Delta U = \frac{8.3}{0.4} \times 8$
$\Delta U = 8.3 \times 20 = 166 \, J$.
Thus,the change in internal energy is $166 \, J$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.
This implies $\frac{T_2}{T_1} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1}$.
Given: Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
Initial volume $V_1 = V$.
Final volume $V_2 = \frac{V}{4}$.
Adiabatic index $\gamma = 1.4$.
Substituting these values into the formula:
$T_2 = T_1 \times \left( \frac{V_1}{V_2} \right)^{\gamma - 1}$
$T_2 = 300 \times \left( \frac{V}{V/4} \right)^{1.4 - 1}$
$T_2 = 300 \times (4)^{0.4} \text{ K}$.

(D) For an adiabatic process,the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.
Since the process is adiabatic,the heat exchange $\Delta Q = 0$.
Therefore,$0 = \Delta U + \Delta W$,which implies $\Delta W = -\Delta U$.
Given that the change in internal energy $\Delta U = -50\, J$.
Substituting the value,we get $\Delta W = -(-50\, J) = +50\, J$.
Thus,the work done during the process is $50\, J$.

(B) The adiabatic modulus of elasticity $(E_{\phi})$ is related to the isothermal modulus of elasticity $(E_{\theta})$ by the adiabatic index $\gamma$ as follows:
$E_{\phi} = \gamma E_{\theta}$
Therefore,the isothermal modulus of elasticity is given by:
$E_{\theta} = \frac{E_{\phi}}{\gamma}$
Given:
$E_{\phi} = 2.1 \times 10^5 \ N/m^2$
$\gamma = \frac{C_p}{C_v} = 1.4$
Substituting the values:
$E_{\theta} = \frac{2.1 \times 10^5}{1.4}$
$E_{\theta} = 1.5 \times 10^5 \ N/m^2$
Thus,the correct option is $B$.