Adiabatic Process Questions in English

(D) thermodynamic process in which there is no exchange of heat between the system and the surroundings is defined as an adiabatic process.
In this process,the heat exchange $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$,so for an adiabatic process,$dU = -dW$.

(A) For an adiabatic process,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 its temperature $(U = n C_v T)$,a decrease in internal energy implies a decrease in temperature $(T)$.
According to the ideal gas equation,$p V = n R T$.
Since $n$ and $R$ are constants and the temperature $T$ decreases during adiabatic expansion,the product $p V$ must also decrease.

(D) $(i)$ Curve $OA$ represents an isobaric process (since pressure is constant).
(ii) Curve $OB$ represents an isothermal process.
(iii) Curve $OC$ represents an adiabatic process,as the slope of an adiabatic process is steeper than that of an isothermal process.
(iv) Curve $OD$ represents an isochoric process (since volume is constant).

(B) The equation for an adiabatic process is given by $P V^{\gamma} = \text{constant}$.
Given that the cube of the pressure is inversely proportional to the fourth power of the volume, we have $P^3 \propto V^{-4}$, which implies $P^3 V^4 = \text{constant}$.
Taking the cube root of both sides, we get $(P^3 V^4)^{1/3} = \text{constant}^{1/3}$, which simplifies to $P V^{4/3} = \text{constant}$.
Comparing this with the standard adiabatic equation $P V^{\gamma} = \text{constant}$, we identify $\gamma = 4/3$.
Thus, the ratio of specific heats is $\gamma = 1.33$.

(B) When a cycle tyre bursts suddenly,the process is adiabatic.
This is because the expansion of air occurs very rapidly,leaving no time for heat exchange between the system (the air inside the tyre) and the surroundings.
Since $dQ = 0$,the process satisfies the condition for an adiabatic process.

(B) For an adiabatic process,the relation between pressure $P$ and temperature $T$ is given by $P^{1-\gamma} T^{\gamma} = \text{constant}$.
Given that $P^2 \propto T^3$,we can write $P \propto T^{3/2}$,which implies $P T^{-3/2} = \text{constant}$.
Comparing $P T^{-3/2} = \text{constant}$ with $P^{1-\gamma} T^{\gamma} = \text{constant}$,we rewrite the given relation as $P T^{-3/2} = \text{constant}$,or $P^{1} T^{-3/2} = \text{constant}$.
Raising to the power of $\frac{1}{1-\gamma}$,we get $P T^{\frac{-3/2}{1-\gamma}} = \text{constant}$.
Comparing the exponents of $T$,we have $\frac{\gamma}{1-\gamma} = -\frac{3}{2}$.
$2\gamma = -3(1-\gamma) \implies 2\gamma = -3 + 3\gamma \implies \gamma = 3$.
For an ideal gas,the molar specific heat at constant volume is $C_V = \frac{R}{\gamma - 1}$.
Substituting $\gamma = 3$,we get $C_V = \frac{R}{3-1} = \frac{R}{2}$.

(A) In an adiabatic compression,work is done on the gas,which increases its internal energy. Since the internal energy of an ideal gas is directly proportional to its absolute temperature $(U \propto T)$,the temperature of the gas increases.
According to the kinetic theory of gases,the average kinetic energy of gas molecules is given by $K_{avg} = \frac{3}{2} k_B T$. As the temperature $T$ increases,the average kinetic energy also increases.
This increase in kinetic energy occurs because,during compression,the molecules collide with the moving piston (wall). When a molecule hits a wall moving towards it,the molecule gains speed (and thus kinetic energy) due to the elastic collision,similar to a ball bouncing off a moving bat. Therefore,both the assertion and the reason are true,and the reason correctly explains the assertion.

(A) The molar specific heat capacity $C$ is defined as the amount of heat required to raise the temperature of $1 \text{ mole}$ of a gas by $1 \text{ K}$.
Mathematically,$C = \frac{\Delta Q}{n \Delta T}$.
In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat is exchanged with the environment.
Therefore,the heat change $\Delta Q = 0$.
Substituting this into the formula,we get $C = \frac{0}{n \Delta T} = 0$.
Thus,the specific heat of a gas undergoing an adiabatic change is $0$.

(B) sudden compression of a gas is an adiabatic process.
For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P_1 V_1^\gamma = P_2 V_2^\gamma$,where $\gamma$ is the adiabatic index (ratio of specific heat capacities).
Given:
Initial pressure $P_1 = P$
Initial volume $V_1 = 400 \ cc$
Final volume $V_2 = 100 \ cc$
Adiabatic index $\gamma = 1.5 = 3/2$
Substituting the values into the adiabatic equation:
$P \times (400)^{3/2} = P_2 \times (100)^{3/2}$
$P_2 = P \times (400/100)^{3/2}$
$P_2 = P \times (4)^{3/2}$
$P_2 = P \times (2^2)^{3/2}$
$P_2 = P \times 2^3$
$P_2 = 8P$
Therefore,the final pressure is $8P$.

(A) For an adiabatic process,the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$,so $W = -\Delta U$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
For a diatomic gas,the molar heat capacity at constant volume is $C_v = \frac{5}{2} R$.
Given: $n = 3$ moles,$\Delta T = -50^{\circ} C$ (since temperature decreases).
Therefore,$\Delta U = 3 \times (\frac{5}{2} R) \times (-50) = 3 \times 2.5 R \times (-50) = -375 R$.
The work done by the gas is $W = -\Delta U = -(-375 R) = 375 R$.

(C) For an adiabatic process, the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$.
Taking the logarithm on both sides, we get $\log P + \gamma \log V = \text{constant}$, which can be written as $\log P = -\gamma \log V + C$.
This is the equation of a straight line $y = mx + c$, where the slope $m = -\gamma$.
From the graph, the slope is calculated as:
$m = \frac{\log P_2 - \log P_1}{\log V_2 - \log V_1} = \frac{2.20 - 2.48}{1.4 - 1.2} = \frac{-0.28}{0.2} = -1.4$.
Since $m = -\gamma$, we have $-\gamma = -1.4$, which implies $\gamma = 1.4$.
Therefore, the ratio of the specific heat capacities is $1.4$.

(C) 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: $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$.
Therefore,the fractional change in pressure is $\frac{dP}{P} = -\gamma \frac{dV}{V}$.
For a diatomic gas,the adiabatic exponent $\gamma = \frac{7}{5} = 1.4$.
Given the increase in volume $\frac{dV}{V} = 0.5 \% = 0.005$.
Substituting the values: $\frac{dP}{P} = -1.4 \times 0.5 \% = -0.7 \%$.
Thus,the change in pressure is $-0.7 \%$.

(A) For a monoatomic gas,the adiabatic index $\gamma = 5/3$. At $STP$,$1$ mole of gas occupies $22.4$ litres. Therefore,the number of moles $n = 5.6 / 22.4 = 0.25$ moles.
For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $V_1 = 5.6$ $L$,$V_2 = 0.7$ $L$,and $T_1 = 273$ $K$.
$T_2 = T_1 (V_1/V_2)^{\gamma-1} = 273 \times (5.6/0.7)^{5/3-1} = 273 \times (8)^{2/3} = 273 \times 4 = 1092$ $K$.
The work done on the gas in an adiabatic process is $W = -\Delta U = -n C_v (T_2 - T_1)$.
For a monoatomic gas,$C_v = 3R/2$.
$W = -0.25 \times (3R/2) \times (1092 - 273) = -0.375 R \times 819 = -307.125 R$.
The work done on the gas is the magnitude,which is approximately $307 R$.

(A) For an adiabatic process, the work done $W$ is given by the formula: $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
Since $P_i V_i^\gamma = P_f V_f^\gamma$, we have $P_f = P_i (V_i / V_f)^\gamma$.
Given $V_f = 8 V_i$ and for a monatomic gas $\gamma = 5/3$, we get $P_f = P_i (1/8)^{5/3} = P_i / 32$.
Substituting these into the work formula: $W = \frac{P_i V_i - (P_i / 32)(8 V_i)}{5/3 - 1} = \frac{P_i V_i - P_i V_i / 4}{2/3} = \frac{(3/4) P_i V_i}{2/3} = \frac{9}{8} P_i V_i$.
Given $W = 180 \text{ J}$ and $P_i = 100 \times 10^3 \text{ Pa}$, we have $180 = \frac{9}{8} \times 10^5 \times V_i$.
$V_i = \frac{180 \times 8}{9 \times 10^5} = \frac{160}{10^5} = 1.6 \times 10^{-3} \text{ m}^3$.
Converting to $\text{cm}^3$: $1.6 \times 10^{-3} \times 10^6 \text{ cm}^3 = 1600 \text{ cm}^3$.

(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Given: $T_1 = 300 \ K$,$V_2 = 2V_1$,and $\gamma = 1.5$.
Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values: $300 \times (V_1)^{1.5-1} = T_2 \times (2V_1)^{1.5-1}$.
$300 \times (V_1)^{0.5} = T_2 \times (2)^{0.5} \times (V_1)^{0.5}$.
$T_2 = \frac{300}{\sqrt{2}} = \frac{300}{1.414} \approx 212.13 \ K$.
The fall in temperature is $\Delta T = T_1 - T_2 = 300 - 212.13 = 87.87 \ K$.
Rounding to the nearest integer,the fall in temperature is approximately $88 \ K$.

(B) Given: $P_1 = 10^5 \text{ Nm}^{-2}$,$V_1 = 16 \text{ L} = 16 \times 10^{-3} \text{ m}^3$,$V_2 = 2 \text{ L} = 2 \times 10^{-3} \text{ m}^3$,$C_v = \frac{3R}{2}$.
For an ideal gas,$C_p = C_v + R = \frac{3R}{2} + R = \frac{5R}{2}$.
The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.
For an adiabatic process,$P_1 V_1^\gamma = P_2 V_2^\gamma$.
$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 10^5 \left( \frac{16}{2} \right)^{5/3} = 10^5 \times (8)^{5/3} = 10^5 \times (2^3)^{5/3} = 10^5 \times 2^5 = 32 \times 10^5 \text{ Nm}^{-2}$.
Work done by the gas $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
$W = \frac{(10^5 \times 16 \times 10^{-3}) - (32 \times 10^5 \times 2 \times 10^{-3})}{5/3 - 1} = \frac{1600 - 6400}{2/3} = \frac{-4800}{2/3} = -4800 \times \frac{3}{2} = -7200 \text{ J} = -7.2 \text{ kJ}$.
Since the work done by the gas is $-7.2 \text{ kJ}$,the work done on the gas is $7.2 \text{ kJ}$.

(C) For an adiabatic process,the work done is given by $W = -\Delta U = -\frac{f}{2} nR \Delta T = \frac{f}{2} nR (T_i - T_f)$.
For the monatomic gas: $n_1 = 2$,$f_1 = 3$,$T_{i1} = 80^{\circ} C$,$T_{f1} = 50^{\circ} C$.
$W = \frac{3}{2} \times 2 \times R \times (80 - 50) = 3R \times 30 = 90R$.
For the diatomic gas: $n_2 = 3$,$f_2 = 5$,$T_{i2} = 50^{\circ} C$,$T_{f2} = 20^{\circ} C$.
$W' = \frac{5}{2} \times 3 \times R \times (50 - 20) = \frac{15}{2} R \times 30 = 15R \times 15 = 225R$.
Now,find the ratio: $\frac{W'}{W} = \frac{225R}{90R} = \frac{225}{90} = 2.5$.
Therefore,$W' = 2.5 W$.

(D) For a monatomic gas, the degrees of freedom $f=3$.
The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Given $T_1 = 630 \,K$, $V_1 = V$, and $V_2 = 27V$.
Substituting these values:
$630 \times V^{\frac{5}{3}-1} = T_2 \times (27V)^{\frac{5}{3}-1}$
$630 = T_2 \times (27)^{\frac{2}{3}}$
$630 = T_2 \times (3^3)^{\frac{2}{3}}$
$630 = T_2 \times 3^2$
$630 = T_2 \times 9$
$T_2 = \frac{630}{9} = 70 \,K$.

(B) Work done on the system,$W = 166.28 \ J$.
Increase in temperature,$\Delta T = 8^{\circ} C = 8 \ K$.
For an adiabatic process,the work done on the gas is given by $W = \frac{nR\Delta T}{\gamma - 1}$.
Given $n = 1 \ mol$ and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.
Substituting the values: $166.28 = \frac{1 \times 8.314 \times 8}{\gamma - 1}$.
$\gamma - 1 = \frac{66.512}{166.28} = 0.4$.
$\gamma = 1.4$.
Since the adiabatic index $\gamma = 1.4$ corresponds to a diatomic gas,the gas is diatomic.

(B) For an adiabatic process,the relation between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$.
From the ideal gas equation,we know that $PV = nRT$,which implies $V = \frac{nRT}{P}$.
Substituting the expression for $V$ into the adiabatic equation:
$P \left( \frac{nRT}{P} \right)^{\gamma} = \text{constant}$.
Since $n$ and $R$ are constants,we can write:
$P \cdot \frac{T^{\gamma}}{P^{\gamma}} = \text{constant}$.
$P^{1-\gamma} T^{\gamma} = \text{constant}$.

(D) According to the first law of thermodynamics,the heat supplied to a 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 between the system and the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation:
$0 = \Delta U + \Delta W$
Therefore,$\Delta U = -\Delta W$.

(D) An adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings $(dQ = 0)$.
For an ideal gas undergoing a reversible adiabatic process,the relationship between pressure $(P)$ and volume $(V)$ is given by $PV^{\gamma} = \text{constant}$,where $\gamma$ is the adiabatic index (ratio of specific heats $C_p/C_v$).
Using the ideal gas law $PV = nRT$,we can express pressure as $P = \frac{nRT}{V}$.
Substituting this into the adiabatic equation: $(\frac{nRT}{V})V^{\gamma} = \text{constant}$.
Since $nR$ is a constant,we get $T V^{\gamma-1} = \text{constant}$.

(A) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Thus,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,which implies $T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$.
Given: Initial temperature $T_1 = 37^{\circ} C = 310.15 \text{ K}$,adiabatic index $\gamma = 1.5$,and final volume $V_2 = \frac{V_1}{2}$.
Substituting these values: $T_2 = 310.15 \times \left( \frac{V_1}{V_1/2} \right)^{1.5-1} = 310.15 \times (2)^{0.5} = 310.15 \times \sqrt{2}$.
Using $\sqrt{2} \approx 1.414$,we get $T_2 = 310.15 \times 1.414 \approx 438.55 \text{ K}$.
Converting back to Celsius: $T_2(^{\circ} C) = 438.55 - 273.15 = 165.4^{\circ} C$.
Rounding to the nearest option,the final temperature is approximately $165.3^{\circ} C$.

(D) For an adiabatic process,the relationship between pressure $p$ and temperature $T$ is given by $T^\gamma p^{1-\gamma} = \text{constant}$.
Thus,$T_1^\gamma p_1^{1-\gamma} = T_2^\gamma p_2^{1-\gamma}$.
Rearranging for $p_2$,we get $p_2 = p_1 \left(\frac{T_1}{T_2}\right)^{\frac{\gamma}{1-\gamma}}$.
Given values: $p_1 = 4 \text{ atm} = 2^2 \text{ atm}$,$\gamma = 5/3$,$T_1 = 27^{\circ} \text{C} = 300 \text{ K}$,and $T_2 = 327^{\circ} \text{C} = 600 \text{ K}$.
Substituting these values:
$p_2 = 2^2 \left(\frac{300}{600}\right)^{\frac{5/3}{1-5/3}} = 2^2 \left(\frac{1}{2}\right)^{\frac{5/3}{-2/3}} = 2^2 \left(\frac{1}{2}\right)^{-5/2} = 2^2 \times 2^{5/2}$.
$p_2 = 2^{2 + 5/2} = 2^{9/2} \text{ atm}$.

(A) 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 $\Delta U = -\Delta W$.
Given work done by the gas $\Delta W = 83 \,J$, so $\Delta U = -83 \,J$.
The change in internal energy is given by $\Delta U = n C_V \Delta T$.
Given $C_p = \frac{11}{10} R$, we find $C_V = C_p - R = \frac{11}{10} R - R = \frac{1}{10} R$.
Substituting the values: $-83 = 1 \times (\frac{1}{10} \times 8.3) \times (T_f - 125)$.
$-83 = 0.83 \times (T_f - 125)$.
$T_f - 125 = \frac{-83}{0.83} = -100$.
$T_f = 125 - 100 = 25^{\circ} C$.

(D) When a compression process is performed suddenly or instantly,there is not enough time for heat transfer; therefore,the process is an adiabatic compression process.
For an adiabatic process,the relation between pressure and volume is given by $P V^\gamma = \text{constant}$.
Thus,$P_i V_i^\gamma = P_f V_f^\gamma$.
Rearranging for the ratio of final pressure to initial pressure: $\frac{P_f}{P_i} = \left( \frac{V_i}{V_f} \right)^\gamma$.
Given $V_f = \frac{V_i}{4}$ and $\gamma = 1.5 = \frac{3}{2}$,we substitute these values:
$\frac{P_f}{P_i} = \left( \frac{V_i}{V_i / 4} \right)^{1.5} = (4)^{1.5} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,the ratio of final pressure to initial pressure is $8:1$.

(D) In an adiabatic process,the system is thermally isolated from its surroundings,meaning no heat exchange $(dQ = 0)$ occurs between the system and the environment.
To maintain this condition,the container must be a perfect thermal insulator.
$A$ thermos flask is designed specifically to minimize heat transfer through conduction,convection,and radiation,making it the best choice among the given options for an adiabatic process.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat supplied to the system,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
Given that the work done by the system equals the decrease in its internal energy,we have $\Delta W = -\Delta U$,which implies $\Delta U + \Delta W = 0$.
Substituting this into the first law equation,we get $\Delta Q = 0$.
$A$ process in which there is no exchange of heat $(\Delta Q = 0)$ with the surroundings is defined as an adiabatic process.
Therefore,the system must have undergone an adiabatic change.

(C) Given: Number of moles $n = 5$, initial temperature $T_1 = 273 \, K$ (at $STP$), final temperature $T_2 = 673 \, K$, gas constant $R = 8.3 \, J \, mol^{-1} \, K^{-1}$, and adiabatic index $\gamma = 1.4$.
For an ideal gas, the change in internal energy $\Delta U$ is given by the formula: $\Delta U = n C_v \Delta T$.
Since $C_v = \frac{R}{\gamma - 1}$, we substitute this into the equation:
$\Delta U = n \left( \frac{R}{\gamma - 1} \right) (T_2 - T_1)$.
Substituting the given values:
$\Delta U = 5 \times \left( \frac{8.3}{1.4 - 1} \right) \times (673 - 273)$.
$\Delta U = 5 \times \left( \frac{8.3}{0.4} \right) \times 400$.
$\Delta U = 5 \times 8.3 \times 1000$.
$\Delta U = 41500 \, J = 41.5 \, kJ$.
Thus, the increase in internal energy is $41.5 \, kJ$.

(C) For an adiabatic process, the relation between pressure and volume is $pV^{\gamma} = \text{constant}$.
Initially, for part $A$: $p_A = p$, $V_A = 5V$. For part $B$: $p_B = 8p$, $V_B = V$.
When the piston is released, it moves until the pressures in both parts are equal. Let the final pressure be $p_f$ and the final volumes be $V_A'$ and $V_B'$.
Since the total volume is constant, $V_A' + V_B' = 5V + V = 6V$.
For adiabatic expansion/compression: $p(5V)^{\gamma} = p_f(V_A')^{\gamma}$ and $(8p)(V)^{\gamma} = p_f(V_B')^{\gamma}$.
Dividing the two equations: $\frac{p(5V)^{\gamma}}{(8p)V^{\gamma}} = \frac{p_f(V_A')^{\gamma}}{p_f(V_B')^{\gamma}} \implies \frac{5^{\gamma}}{8} = \left(\frac{V_A'}{V_B'}\right)^{\gamma}$.
Given $\gamma = 1.5 = \frac{3}{2}$, we have $\frac{5^{3/2}}{8} = \left(\frac{V_A'}{V_B'}\right)^{3/2}$.
Taking the power of $2/3$ on both sides: $\left(\frac{5^{3/2}}{8}\right)^{2/3} = \frac{V_A'}{V_B'} \implies \frac{5}{8^{2/3}} = \frac{V_A'}{V_B'} \implies \frac{5}{4} = \frac{V_A'}{V_B'}$.
Thus, $V_A' = \frac{5}{4}V_B'$.
Substituting into $V_A' + V_B' = 6V$: $\frac{5}{4}V_B' + V_B' = 6V \implies \frac{9}{4}V_B' = 6V \implies V_B' = \frac{24}{9}V = \frac{8}{3}V$.
Then $V_A' = 6V - \frac{8}{3}V = \frac{18-8}{3}V = \frac{10}{3}V$.

(A) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Given that the process follows $p V^{3/2} = \text{constant}$, we identify the adiabatic index $\gamma = 3/2$.
Let the initial temperature be $T_i = T$ and the initial volume be $V_i = V$.
The gas is compressed to half its initial volume, so $V_f = V/2$.
Using the relation $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$, we get:
$T_f = T_i \left( \frac{V_i}{V_f} \right)^{\gamma-1}$
Substituting the values: $T_f = T \left( \frac{V}{V/2} \right)^{3/2 - 1}$
$T_f = T (2)^{1/2} = \sqrt{2} T$.

(D) The root mean square (rms) speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
Since $v_{rms} \propto \sqrt{T}$,if the rms speed becomes half,the temperature $T$ becomes $(1/2)^2 = 1/4$ of the initial temperature.
So,$T_f = \frac{T_i}{4}$.
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: $T_i (200)^{\gamma-1} = \frac{T_i}{4} (V_f)^{\gamma-1}$.
Given $\gamma = 1.5$,so $\gamma - 1 = 0.5 = 1/2$.
$200^{1/2} = \frac{1}{4} V_f^{1/2}$.
Squaring both sides: $200 = \frac{1}{16} V_f$.
$V_f = 200 \times 16 = 3200 \ cc$.

(D) For an adiabatic process, the first law of thermodynamics states that $Q = \Delta U + W$. Since the process is adiabatic, $Q = 0$, which implies $W = -\Delta U$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
Given: $n = 1 \,mol$, $C_v = 0.8 \,kJ \,kg^{-1} \,K^{-1}$. Assuming the molar mass $M = 0.05 \,kg/mol$ (standard for such problems to yield integer results), $C_{v,molar} = M \times C_v = 0.05 \times 800 \,J \,mol^{-1} \,K^{-1} = 40 \,J \,mol^{-1} \,K^{-1}$.
Then, $\Delta U = 1 \,mol \times 40 \,J \,mol^{-1} \,K^{-1} \times (250 \,K - 200 \,K) = 40 \times 50 = 2000 \,J = 2 \,kJ$.
However, if we use the provided specific heat directly as molar heat capacity $C_v = 0.8 \,kJ \,mol^{-1} \,K^{-1}$ (often implied in such textbook problems), then $\Delta U = 1 \times 0.8 \times (250 - 200) = 0.8 \times 50 = 40 \,kJ$.
Since $W = -\Delta U$, the magnitude of work done is $40 \,kJ$.

(B) For an adiabatic process,the work done is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.
Helium is a monoatomic gas,so $\gamma = 5/3$.
At $STP$,$1 \ mole$ of gas occupies $22.4 \ L$. Given $V_i = 5.6 \ L$,the number of moles $n = \frac{5.6}{22.4} = 0.25 \ mole = \frac{1}{4} \ mole$.
For an adiabatic process,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.
$T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1} = T \left(\frac{5.6}{0.7}\right)^{(5/3)-1} = T(8)^{2/3} = T(2^3)^{2/3} = 4T$.
Now,$W = \frac{n R (T - 4T)}{(5/3) - 1} = \frac{(1/4) R (-3T)}{2/3} = \frac{-3/4 RT}{2/3} = -\frac{9}{8} RT$.
Since the gas is compressed,work is done on the gas,so the work done by the gas is negative.

(B) From the first law of thermodynamics, we have the equation $dQ = dU + dW$.
In an adiabatic process, there is no exchange of heat with the surroundings, so $dQ = 0$.
Substituting this into the equation, we get $0 = dU + dW$, which implies $dU = -dW$.
Given that the gas does $4.5 \,J$ of external work, $dW = 4.5 \,J$.
Therefore, $dU = -4.5 \,J$.
Since the change in internal energy $dU$ is negative, the internal energy decreases by $4.5 \,J$.

(B) sudden compression is an adiabatic process. The relation between temperature and volume for an adiabatic process is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Initial temperature $T_1 = 127 + 273 = 400 \ K$.
Given $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.
Using the adiabatic relation:
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}$
$T_2 = 400 \times \left( \frac{27}{8} \right)^{\frac{5}{3}-1} = 400 \times \left( \frac{27}{8} \right)^{\frac{2}{3}}$
$T_2 = 400 \times \left( \left( \frac{3}{2} \right)^3 \right)^{\frac{2}{3}} = 400 \times \left( \frac{3}{2} \right)^2 = 400 \times \frac{9}{4} = 900 \ K$.
The rise in temperature is $\Delta T = T_2 - T_1 = 900 \ K - 400 \ K = 500 \ K$.

(B) The compression ratio is given by $r = V_1/V_2 = 20$.
Thus,$V_2 = V_1/20 = (10^{-3}/20) \ m^3$.
For an adiabatic process,$p_1 V_1^\gamma = p_2 V_2^\gamma$.
Therefore,$p_2 = p_1 (V_1/V_2)^\gamma = 10^5 \times (20)^{1.4} = 10^5 \times 66.3 = 66.3 \times 10^5 \ Pa$.
The work done by the gas during an adiabatic process is given by $W = \frac{p_1 V_1 - p_2 V_2}{\gamma - 1}$.
Substituting the values: $W = \frac{(10^5 \times 10^{-3}) - (66.3 \times 10^5 \times 10^{-3} / 20)}{1.4 - 1}$.
$W = \frac{100 - 331.5}{0.4} = \frac{-231.5}{0.4} = -578.75 \ J \approx -579 \ J$.
Since the work is done on the gas,the value is negative.

(B) For an adiabatic process, the relation between pressure $P$ and volume $V$ is given by $P V^\gamma = \text{constant}$.
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 $\frac{dV}{V} = -\frac{1}{\gamma} \frac{dP}{P}$.
Given that the pressure is reduced by $0.1 \%$, we have $\frac{dP}{P} = -0.001$.
Given $\gamma = \frac{5}{3}$, we substitute these values into the equation:
$\frac{dV}{V} = -\frac{1}{5/3} \times (-0.001) = \frac{3}{5} \times 0.001 = 0.6 \times 0.001 = 0.0006$.
Converting this to percentage: $0.0006 \times 100 \% = 0.06 \%$.
Thus, the volume increases by $0.06 \%$.

(A) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Given,initial volume $V_1 = V$,final volume $V_2 = 3V$,and initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K}$.
The adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$.
Using the relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,we get:
$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$
$\frac{T_2}{300} = \left(\frac{V}{3V}\right)^{\frac{5}{3}-1} = \left(\frac{1}{3}\right)^{\frac{2}{3}}$
$T_2 = 300 \times (3)^{-\frac{2}{3}} = 300 \times \frac{1}{3^{0.666}} \approx 300 \times 0.4807 \approx 144.2 \text{ K}$.

(A) Given: Number of moles $n = 2$,Initial temperature $T_1 = 327 + 273 = 600 \ K$,Ratio of specific heats $\gamma = \frac{4}{3}$.
Volume increases by $700 \%$,so $V_2 = V_1 + 7V_1 = 8V_1$.
For an adiabatic process,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
$T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = 600 \left( \frac{1}{8} \right)^{\frac{4}{3}-1} = 600 \left( \frac{1}{8} \right)^{1/3} = 600 \times \frac{1}{2} = 300 \ K$.
Work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
$W = \frac{2 \times 8.3 \times (600 - 300)}{\frac{4}{3} - 1} = \frac{2 \times 8.3 \times 300}{1/3} = 2 \times 8.3 \times 300 \times 3 = 14940 \ J = 14.94 \ kJ$.

(D) For a sudden compression,the process is adiabatic.
For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Here,$\gamma = 1.5$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given $T_2 = 2T_1$ and $\gamma = 1.5$.
Using the relation $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$,we get:
$T_1 V_1^{1.5-1} = 2T_1 V_2^{1.5-1}$
$V_1^{0.5} = 2 V_2^{0.5}$
Squaring both sides: $V_1 = 4 V_2$,which implies $V_2 = \frac{V_1}{4} = 0.25 V_1$.
The decrease in volume is $\Delta V = V_1 - V_2 = V_1 - 0.25 V_1 = 0.75 V_1$.
The percentage decrease is $\frac{\Delta V}{V_1} \times 100 = 0.75 \times 100 = 75\%$.

(B) According to the first law of thermodynamics,the change in internal energy $dU$ is given by $dU = dQ - dW$,where $dQ$ is the heat supplied to the system and $dW$ is the work done by the system.
Given that the work done by the system is equal to the decrease in its internal energy,we have $dW = -dU$.
Substituting this into the first law equation:
$dU = dQ - (-dU)$
$dU = dQ + dU$
$dQ = 0$
Since the heat exchange $dQ$ is zero,the process is an adiabatic process.

(B) For an adiabatic process,the relation between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the given values: $T V^{\gamma-1} = (T/2) (4V)^{\gamma-1}$.
Dividing both sides by $T V^{\gamma-1}$,we get $1 = (1/2) (4)^{\gamma-1}$,which implies $2 = 4^{\gamma-1}$.
Since $4 = 2^2$,we have $2^1 = (2^2)^{\gamma-1} = 2^{2\gamma-2}$.
Equating the exponents: $1 = 2\gamma - 2$,so $2\gamma = 3$,which gives $\gamma = 3/2$.
The work done in an adiabatic process is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.
Substituting $T_1 = T$,$T_2 = T/2$,and $\gamma = 3/2$: $W = \frac{nR(T - T/2)}{3/2 - 1} = \frac{nR(T/2)}{1/2} = nRT$.
Since $PV = nRT$ for an ideal gas,$W = PV$.
Comparing $W = \alpha PV$ with $W = PV$,we get $\alpha = 1$.

(C) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Let the initial temperature be $T_1 = T$ and the initial volume be $V_1 = V$.
According to the problem,the final volume is $V_2 = 2V$ and the final temperature is $T_2 = T/2$.
Substituting these values into the adiabatic relation:
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$
$T \cdot V^{\gamma-1} = (T/2) \cdot (2V)^{\gamma-1}$
Dividing both sides by $T \cdot V^{\gamma-1}$:
$1 = (1/2) \cdot 2^{\gamma-1}$
$2 = 2^{\gamma-1}$
Since the bases are equal,the exponents must be equal:
$1 = \gamma - 1$
$\gamma = 2$.

(B) The process of a tyre bursting is an adiabatic expansion process.
For an adiabatic process,the relation between temperature $T$ and pressure $P$ is given by $T^\gamma P^{1-\gamma} = \text{constant}$.
Thus,$T_1^\gamma P_1^{1-\gamma} = T_2^\gamma P_2^{1-\gamma}$.
Given: $P_1 = 2 \text{ atm}$,$P_2 = 1 \text{ atm}$ (atmospheric pressure),$T_1 = T$,and $\gamma = \frac{3}{2}$.
Substituting these values:
$T^{\frac{3}{2}} (2)^{1 - \frac{3}{2}} = T_2^{\frac{3}{2}} (1)^{1 - \frac{3}{2}}$
$T^{\frac{3}{2}} (2)^{-\frac{1}{2}} = T_2^{\frac{3}{2}}$
$T_2 = T \times (2)^{-\frac{1}{2} \times \frac{2}{3}}$
$T_2 = T \times (2)^{-\frac{1}{3}}$
$T_2 = \left(\frac{1}{2}\right)^{1/3} T$.

(B) process in which there is no heat exchange $(Q = 0)$ is called an adiabatic process.
For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
Squaring both sides,we get $T^2 V^{2(\gamma-1)} = \text{constant}$.
Comparing this with the given law $T^2 V^\alpha = \text{constant}$,we find $\alpha = 2(\gamma-1)$.
For a polyatomic gas,the adiabatic index $\gamma$ is typically $\frac{4}{3}$.
Substituting $\gamma = \frac{4}{3}$ into the expression for $\alpha$:
$\alpha = 2 \left( \frac{4}{3} - 1 \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$.

(B) For an adiabatic process,the relation between pressure $p$ and volume $V$ is given by $p V^\gamma = \text{constant}$.
For the adiabatic process $A \rightarrow B$,the states are $(2p_0, V_0)$ and $(p_0, V_1)$.
Thus,$(2p_0) V_0^\gamma = p_0 V_1^\gamma$.
Dividing by $p_0$,we get $2 V_0^\gamma = V_1^\gamma$,which implies $V_1 = 2^{1/\gamma} V_0$.
The work done $W$ in an adiabatic process is given by $W = \frac{p_i V_i - p_f V_f}{\gamma - 1}$.
Substituting the values for process $A \rightarrow B$:
$W = \frac{(2p_0)(V_0) - (p_0)(V_1)}{\gamma - 1} = \frac{2p_0 V_0 - p_0 (2^{1/\gamma} V_0)}{\gamma - 1}$.
Rearranging the terms,we get $W = \frac{p_0 V_0 (2 - 2^{1/\gamma})}{\gamma - 1} = \frac{p_0 V_0 (2^{1/\gamma} - 2)}{1 - \gamma}$.

(B) The work done in an adiabatic process is given by the formula: $\Delta W = \frac{\mu R \Delta T}{\gamma - 1}$.
For $1$ mole of monoatomic gas $(\mu = 1)$: $\Delta T = 20^{\circ} C$,$\gamma = 5/3$. Therefore,$W = \frac{1 \times R \times 20}{(5/3 - 1)} = \frac{20R}{2/3} = 30R$ ... $(i)$.
For $6$ moles of rigid diatomic gas $(\mu = 6)$: $\Delta T = 20^{\circ} C$,$\gamma = 7/5$. The work done $W'$ is given by: $W' = \frac{6 \times R \times 20}{(7/5 - 1)} = \frac{120R}{2/5} = \frac{120R \times 5}{2} = 300R$.
Comparing $W'$ with $W$: $W' = 300R = 10 \times (30R) = 10W$.

(D) An ideal-gas process in which no heat is exchanged between the system and its surroundings is defined as an adiabatic process.
In this process,the heat transfer $dQ = 0$.
According to the first law of thermodynamics,$dU = dQ - dW$,which simplifies to $dU = -dW$ for an adiabatic process.

(D) Given,initial temperature of $1$ mole of $N_2$ gas,$T_1 = 300 \,K$. Initial pressure of gas,$p_1 = p$. Final pressure of gas,$p_2 = 10p$. For a rigid diatomic gas,the adiabatic index $\gamma = 7/5 = 1.4$.
Using the adiabatic relation between pressure and temperature: $T_1^\gamma p_1^{1-\gamma} = T_2^\gamma p_2^{1-\gamma}$.
Rearranging for $T_2$: $(T_2/T_1)^\gamma = (p_1/p_2)^{1-\gamma} = (p_2/p_1)^{\gamma-1}$.
$T_2 = T_1 \times (p_2/p_1)^{(\gamma-1)/\gamma} = 300 \times (10p/p)^{(1.4-1)/1.4} = 300 \times 10^{2/7}$.
Since $10^{2/7} = (10^2)^{1/7} = 100^{1/7} = 1.9$,we have $T_2 = 300 \times 1.9 = 570 \,K$.