Adiabatic Process Questions in English

(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 work is done on the gas,the work done by the gas is $\Delta W = -45 \,J$.
Substituting these values into the first law equation:
$0 = \Delta U + (-45 \,J)$
$\Delta U = 45 \,J$.
Thus,the change in internal energy is $45 \,J$ and the heat flowed into the gas is $0$.

(B) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = \text{constant}$.
For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.
Given that the gas is compressed to $1/8$ of its initial volume,we have $V_1 = V$ and $V_2 = \frac{V}{8}$.
Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values: $T_1 V^{\frac{5}{3}-1} = T_2 \left(\frac{V}{8}\right)^{\frac{5}{3}-1}$.
$T_1 V^{\frac{2}{3}} = T_2 \left(\frac{V}{8}\right)^{\frac{2}{3}}$.
$\frac{T_2}{T_1} = \left(\frac{V}{V/8}\right)^{\frac{2}{3}} = 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$.
The mean kinetic energy $\langle E \rangle$ of a gas molecule is given by $\langle E \rangle = \frac{3}{2} k_B T$,which implies $\langle E \rangle \propto T$.
Therefore,$\frac{\langle E_2 \rangle}{\langle E_1 \rangle} = \frac{T_2}{T_1} = 4$.

(C) The change in internal energy $dU$ for an ideal gas is given by the formula:
$dU = n C_v dT$
Since $C_v = \frac{R}{\gamma - 1}$, the formula becomes:
$dU = n \frac{R}{\gamma - 1} (T_2 - T_1)$
Given values:
$n = 5 \, mol$
$T_1 = 273 \, K$ (at $STP$)
$T_2 = 673 \, K$
$R = 8.3 \, J/mol-K$
$\gamma = 1.4$
Substituting these values into the equation:
$dU = 5 \times \frac{8.3}{1.4 - 1} \times (673 - 273)$
$dU = 5 \times \frac{8.3}{0.4} \times 400$
$dU = 5 \times 8.3 \times 1000$
$dU = 41500 \, J$
Converting to kilo joules:
$dU = 41.50 \, kJ$

(D) For an adiabatic process, the relation between pressure $p$ and volume $V$ is given by $pV^\gamma = \text{constant}$, where $\gamma = C_p / C_V$.
Given that $p \propto T^3$, we can write $p = k T^3$.
Using the ideal gas equation $pV = nRT$, we have $T = \frac{pV}{nR}$.
Substituting this into the given relation:
$p = k \left( \frac{pV}{nR} \right)^3$
$p = k \frac{p^3 V^3}{(nR)^3}$
$1 = \left( \frac{k}{(nR)^3} \right) p^2 V^3$
$p^2 V^3 = \text{constant}'$
$p V^{3/2} = \text{constant}''$
Comparing this with the standard adiabatic equation $pV^\gamma = \text{constant}$, we get $\gamma = 3/2$.
Thus, the value of $C_p / C_V$ is $3/2$.

(A) For an adiabatic process,the equation of state is given by $p V^\gamma = \text{constant}$.
Using the ideal gas law $pV = nRT$,we can substitute $V = \frac{nRT}{p}$ into the adiabatic equation:
$p \left( \frac{nRT}{p} \right)^\gamma = \text{constant}$
$p \cdot p^{-\gamma} \cdot T^\gamma = \text{constant}$
$p^{1-\gamma} T^\gamma = \text{constant}$.
Thus,option $A$ is correct.

(D) For an adiabatic process,the relation between pressure $P$ and volume $V$ is given by $PV^{\gamma} = \text{constant}$.
As the volume $V$ decreases during compression,the pressure $P$ must increase to keep the product constant.
According to the ideal gas law $PV = nRT$,since $P$ increases and $V$ decreases,the temperature $T$ must increase because the work done on the gas increases its internal energy in an adiabatic process ($Q = 0$,$\Delta U = -W$).
Therefore,in adiabatic compression,both temperature and pressure increase.

(B) In an adiabatic process,the system does not exchange heat with the surroundings,so $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q - W$.
Since $Q = 0$,we have $\Delta U = -W$.
In an adiabatic expansion,the gas does work on the surroundings,so $W > 0$.
This implies $\Delta U < 0$,which means the internal energy of the system 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 system.

(D) For an adiabatic process,the relationship between pressure $P$ and density $\rho$ (or $d$) is given by $P \propto \rho^\gamma$.
Thus,$\frac{P^{\prime}}{P} = \left(\frac{d^{\prime}}{d}\right)^\gamma$.
Given $\frac{d^{\prime}}{d} = 32$ and $\gamma = \frac{7}{5}$.
Substituting these values: $\frac{P^{\prime}}{P} = (32)^{7/5}$.
Since $32 = 2^5$,we have $\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$.
Calculating $2^7 = 128$.
Therefore,the ratio $\frac{P^{\prime}}{P} = 128$.

(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 adiabatic process,the molar heat capacity at constant volume is $C_v = \frac{R}{\gamma - 1}$.
Given: $n = 5 \ mol$,$\gamma = \frac{7}{5}$,$R = 8.30 \ J \ mol^{-1} \ K^{-1}$.
Initial temperature $T_1 = 0^{\circ} C = 273 \ K$.
Final temperature $T_2 = 400^{\circ} C = 673 \ K$.
Change in temperature $\Delta T = T_2 - T_1 = 400 \ K$.
Substituting the values:
$\Delta U = n \left( \frac{R}{\gamma - 1} \right) \Delta T$
$\Delta U = 5 \times \left( \frac{8.30}{\frac{7}{5} - 1} \right) \times 400$
$\Delta U = 5 \times \left( \frac{8.30}{2/5} \right) \times 400$
$\Delta U = 5 \times \left( \frac{8.30 \times 5}{2} \right) \times 400$
$\Delta U = 5 \times 20.75 \times 400 = 41500 \ J$.
Converting to kilo-joules: $\Delta U = 41.50 \ kJ$.
Rounding to the nearest provided option,the increase in internal energy is $41.55 \ kJ$.

(C) For an adiabatic process,the relationship between pressure $P$ and density $d$ is given by $P \propto d^\gamma$.
Given $\gamma = \frac{7}{5}$ and $\frac{d^{\prime}}{d} = 32$.
We have the relation $\frac{P^{\prime}}{P} = \left(\frac{d^{\prime}}{d}\right)^\gamma$.
Substituting the given values:
$\frac{P^{\prime}}{P} = (32)^{7/5}$.
Since $32 = 2^5$,we have:
$\frac{P^{\prime}}{P} = (2^5)^{7/5} = 2^7$.
Calculating the value: $2^7 = 128$.
Therefore,the ratio $\frac{P^{\prime}}{P}$ is $128$.

(C) For an adiabatic process,the work done is given by the formula: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
For a diatomic gas,the adiabatic index $\gamma = 1.4$.
Given: $P_1 = 2 \times 10^5 \ N \ m^{-2}$,$V_1 = 2 \ m^3$,$V_2 = 0.5 \ m^3$.
Using the adiabatic relation $P_1 V_1^\gamma = P_2 V_2^\gamma$,we find $P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma = 2 \times 10^5 \times \left(\frac{2}{0.5}\right)^{1.4} = 2 \times 10^5 \times (4)^{1.4}$.
Given $(4)^{1.4} = 6.96$,so $P_2 = 2 \times 10^5 \times 6.96 = 13.92 \times 10^5 \ N \ m^{-2}$.
Now,substitute the values into the work formula:
$W = \frac{(2 \times 10^5 \times 2) - (13.92 \times 10^5 \times 0.5)}{1.4 - 1}$
$W = \frac{4 \times 10^5 - 6.96 \times 10^5}{0.4} = \frac{-2.96 \times 10^5}{0.4} = -7.4 \times 10^5 \ J$.
The negative sign indicates that work is done on the gas during compression.

(B) Given the internal energy $U = A P^2 V$.
For an adiabatic process, the first law of thermodynamics states $dQ = dU + dW = 0$, which implies $dU = -dW = -P dV$.
Thus, $dU = -P dV$.
Taking the differential of $U$ with respect to $V$: $dU = A P^2 dV + 2 A P V dP$.
Equating the two expressions for $dU$: $-P dV = A P^2 dV + 2 A P V dP$.
Dividing by $P$ (assuming $P \neq 0$): $-dV = A P dV + 2 A V dP$.
Rearranging the terms: $-dV - A P dV = 2 A V dP \Rightarrow -(1 + A P) dV = 2 A V dP$.
Separating the variables: $-\frac{dV}{V} = \frac{2 A dP}{1 + A P}$.
Integrating both sides: $-\int \frac{dV}{V} = \int \frac{2 A dP}{1 + A P}$.
$-\ln V = 2 \ln(1 + A P) + C$.
This simplifies to $\ln V + \ln(1 + A P)^2 = \text{constant}$.
Therefore, $V(1 + A P)^2 = \text{constant}$.

(B) For an adiabatic process,the relationship between temperature and volume is given by $T V^{\gamma-1} = \text{constant}$.
Let the temperature after adiabatic compression be $T$. Then,$T_{0} V_{0}^{\gamma-1} = T \left(\frac{V_{0}}{2}\right)^{\gamma-1}$.
Solving for $T$,we get $T = T_{0} 2^{\gamma-1}$.
When the gas is allowed to come to thermal equilibrium with the surroundings at constant volume $\frac{V_{0}}{2}$,the heat released $\Delta Q$ is equal to the change in internal energy: $\Delta Q = n C_{V} \Delta T$.
Using $C_{V} = \frac{R}{\gamma-1}$ and the ideal gas law $n R T_{0} = p_{0} V_{0}$,we have:
$\Delta Q = n \left(\frac{R}{\gamma-1}\right) (T - T_{0}) = \frac{n R T_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.
Substituting $n R T_{0} = p_{0} V_{0}$,the heat released is $\frac{p_{0} V_{0}}{\gamma-1} (2^{\gamma-1} - 1)$.

(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = \text{constant}$.
Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.
Given: $V_2 = 8V_1$ and $T_2 = T_1/4$.
Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.
Substituting the values: $T_1 V_1^{\gamma-1} = (T_1/4) (8V_1)^{\gamma-1}$.
Dividing both sides by $T_1 V_1^{\gamma-1}$: $1 = (1/4) \cdot 8^{\gamma-1}$.
$4 = 8^{\gamma-1}$.
Expressing in base $2$: $2^2 = (2^3)^{\gamma-1} = 2^{3\gamma-3}$.
Equating the exponents: $2 = 3\gamma - 3$.
$3\gamma = 5$, which gives $\gamma = 5/3$.
A gas with adiabatic index $\gamma = 5/3$ is a monoatomic gas.
Among the given options, Helium (He) is a monoatomic gas. Therefore, the correct option is $D$.

(B) The right side compartment undergoes an adiabatic compression because the piston is adiabatic and frictionless.
For an adiabatic process,the relation is $PV^{\gamma} = \text{constant}$.
Given $\gamma = 1.5 = 3/2$.
Let the initial state be $(P, V_1 = 9)$ and the final state be $(P_2 = \frac{27}{8}P, V_2)$.
Since the piston is frictionless,the pressure on both sides must be equal at equilibrium. Thus,the final pressure on the right side is also $\frac{27}{8}P$.
Using the adiabatic relation: $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$.
$P(9)^{3/2} = \frac{27}{8}P(V_2)^{3/2}$.
$(9)^{3/2} = \frac{27}{8} V_2^{3/2}$.
$27 = \frac{27}{8} V_2^{3/2}$.
$V_2^{3/2} = 8$.
$V_2 = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4 \text{ litres}$.

(C) For an adiabatic process,$PV^\gamma = \text{constant}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$.
Given initial state: $P_1 = P, V_1 = V$.
Final volume: $V_2 = 27V$.
Using the adiabatic relation $P_1V_1^\gamma = P_2V_2^\gamma$,we find the final pressure $P_2$:
$P_2 = P(V_1/V_2)^\gamma = P(V/27V)^{5/3} = P(1/27)^{5/3} = P(1/3^3)^{5/3} = P(1/3^5) = P/243$.
The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = \frac{f}{2} (P_2V_2 - P_1V_1)$,where $f$ is the degrees of freedom.
For a monoatomic gas,$f = 3$.
Substituting the values:
$\Delta U = \frac{3}{2} ((\frac{P}{243}) \cdot 27V - PV) = \frac{3}{2} (\frac{P}{9} - PV) = \frac{3}{2} (\frac{P - 9P}{9}) = \frac{3}{2} (\frac{-8P}{9}) = -\frac{4}{3}PV$.

(B) For an adiabatic process,the adiabatic index $\gamma = c_p / c_v = 3R / 2R = 1.5$.
Initial state: $P_i = 8 \ \text{bar} = 8 \times 10^5 \ \text{Pa}$,$V_i = 0.15 \ \text{m}^3$.
Final state: $P_f = 1 \ \text{bar} = 1 \times 10^5 \ \text{Pa}$.
Using the adiabatic relation $P_i V_i^\gamma = P_f V_f^\gamma$,we find the final volume $V_f$:
$V_f = V_i (P_i / P_f)^{1/\gamma} = 0.15 \times (8/1)^{1/1.5} = 0.15 \times (8)^{2/3} = 0.15 \times 4 = 0.6 \ \text{m}^3$.
The work done in an adiabatic process is given by $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
$W = \frac{(8 \times 10^5 \times 0.15) - (1 \times 10^5 \times 0.6)}{1.5 - 1} = \frac{120000 - 60000}{0.5} = \frac{60000}{0.5} = 120000 \ \text{J} = 120 \ \text{kJ}$.