First law of thermodynamics Questions in English

(D) For an adiabatic process,the heat exchange $q = 0$.
In a free expansion process,the gas expands against zero external pressure,so the work done $w = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$ and $w = 0$,the change in internal energy $\Delta U = 0 + 0 = 0 \ J$.

(C) Since the system absorbs heat,$q = +208\, J$.
For a reversible isothermal expansion,the work done $w$ is given by the formula: $w = -nRT \ln(\frac{V_2}{V_1})$.
Given: $n = 0.04\, mol$,$R = 8.314\, J / mol\, K$,$T = 37 + 273 = 310\, K$,$V_1 = 50.0\, mL$,$V_2 = 375\, mL$.
Ratio $\frac{V_2}{V_1} = \frac{375}{50} = 7.5$.
Substituting the values: $w = -0.04 \times 8.314 \times 310 \times \ln(7.5)$.
$w = -0.04 \times 8.314 \times 310 \times 2.01 = -207.22\, J \approx -208\, J$.
Thus,$q = +208\, J$ and $w = -208\, J$.

(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species,which is $1 - 0 = 1$.
The temperature $T = 25 + 273 = 298 \ K$.
The gas constant $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values: $82.8 \ kJ = \Delta U + (1 \times 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1} \times 298 \ K)$.
$82.8 \ kJ = \Delta U + 2.477 \ kJ$.
$\Delta U = 82.8 - 2.477 = 80.323 \ kJ \approx 80.32 \ kJ$.

(B) The relationship between $\Delta H$ and $\Delta E$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$.
First,calculate the change in the number of gaseous moles,$\Delta n_{g}$:
$\Delta n_{g} = (3 + 3) - (1 + 3) = 6 - 4 = 2$.
Given:
$\Delta E = 17 \ kcal$,
$R = 2 \times 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$,
$T = 27 + 273 = 300 \ K$.
Substituting the values into the equation:
$\Delta H = 17 + (2 \times 2 \times 10^{-3} \times 300) = 17 + 1.2 = 18.2 \ kcal$.

(A) For an isothermal process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $\Delta U = 0$,we have $q = -w$.
For a reversible expansion process,the work done is $w = -2.303 \ nRT \ \log \ \frac{V_2}{V_1}$.
Therefore,$q = -(-2.303 \ nRT \ \log \ \frac{V_2}{V_1}) = 2.303 \ nRT \ \log \ \frac{V_2}{V_1}$.

(B) For an adiabatic process,the work done is given by the change in internal energy: $W = \Delta U = nC_v(T_2 - T_1)$.
Given: $n = 1\, mol$,$C_v = 20\, J/K$,$T_1 = 27 + 273 = 300\, K$,and $W = -3000\, J$ (work done by the system is negative).
Substituting the values: $-3000 = 1 \times 20 \times (T_2 - 300)$.
$-150 = T_2 - 300$.
$T_2 = 300 - 150 = 150\, K$.

(A) The container is insulated,which means no heat exchange occurs with the surroundings. Therefore,the process is adiabatic,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$.
Since $q = 0$,we have $\Delta U = W$.
Stirring the liquid with a paddle performs work on the system,so $W \neq 0$,which consequently leads to $\Delta U \neq 0$.

(C) According to the first law of thermodynamics,$\Delta U = q + w$.
For an adiabatic process,the heat exchange $q = 0$.
Therefore,$\Delta U = w$.
If work is done by the system,$w$ is negative,so $\Delta U = -w_{by}$,which means the internal energy decreases by an amount equal to the work done by the system.
Thus,option $C$ is correct.

(B) The balanced chemical equation for the combustion of methane is:
$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$
Calculate the change in the number of gaseous moles,$\Delta n_g$:
$\Delta n_g = (n_{products, gas}) - (n_{reactants, gas}) = 1 - (1 + 2) = 1 - 3 = -2$
Using the relation between enthalpy change and internal energy change:
$\Delta H^o = \Delta U^o + \Delta n_g RT$
Substituting the value of $\Delta n_g$:
$\Delta H^o = \Delta U^o + (-2)RT$
$\Delta H^o = \Delta U^o - 2RT$
Since $2RT$ is a positive value,subtracting it from $\Delta U^o$ results in a value smaller than $\Delta U^o$:
$\therefore \Delta H^o < \Delta U^o$

(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + W$. Since $q = 0$,$\Delta U = W$.
Given that the gas performs work,$W = -75 \, cal$.
The change in internal energy for a monoatomic ideal gas is $\Delta U = n C_v \Delta T$.
For a monoatomic gas,$C_v = \frac{3}{2} R$.
Initial temperature $T_1 = 227 + 273 = 500 \, K$.
Substituting the values: $-75 = 0.10 \times (\frac{3}{2} \times 2) \times (T_2 - 500)$.
$-75 = 0.10 \times 3 \times (T_2 - 500)$.
$-75 = 0.3 \times (T_2 - 500)$.
$-250 = T_2 - 500$.
$T_2 = 500 - 250 = 250 \, K$.

(C) The work done during expansion against a variable pressure is given by $W = -\int_{V_1}^{V_2} P \, dV$.
Substituting $P = \frac{20}{V}$,we get $W = -\int_{1}^{10} \frac{20}{V} \, dV$.
$W = -20 \ln(\frac{10}{1}) = -20 \times 2.303 \log_{10}(10) = -46.06 \ L \cdot atm$.
Converting work to Joules: $W = -46.06 \times 101.3 \ J \approx -4665.88 \ J$.
According to the first law of thermodynamics,$\Delta U = q + W$.
Given $\Delta U = 400 \ J$,we have $400 = q - 4665.88$.
Therefore,$q = 400 + 4665.88 = 5065.88 \ J$.

(A) The work done during expansion is given by $w = -P_{ext} \Delta V$.
For the vaporisation of $1 \ mole$ of water: $H_2O(l) \rightarrow H_2O(g)$.
Volume of $1 \ mole$ of liquid water $(V_1)$: Since density $\rho \approx 1 \ g/mL$,$V_1 = \frac{18 \ g}{1 \ g/mL} = 18 \ mL = 0.018 \ L$.
Volume of $1 \ mole$ of water vapour $(V_2)$ assuming ideal gas behavior: $V_2 = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 373}{1} = 30.62 \ L$.
Change in volume $\Delta V = V_2 - V_1 = 30.62 - 0.018 \approx 30.6 \ L$.
Work done $w = -P \Delta V = -1 \ atm \times 30.6 \ L = -30.6 \ L \ atm$.
Converting to Joules: $w = -30.6 \times 101.3 \ J \approx -3099.78 \ J \approx -3100 \ J$.

(B) For an isothermal irreversible expansion of an ideal gas,the work done is given by: $w = -P_{ext} \Delta V$.
Since $V = \frac{nRT}{P}$,we have $\Delta V = nRT (\frac{1}{P_2} - \frac{1}{P_1})$.
Given $P_{ext} = P_2 = 1 \, atm$,$n = 5 \, mol$,$T = 300 \, K$,$P_1 = 10 \, atm$,and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values: $w = -nRT P_{ext} (\frac{1}{P_2} - \frac{1}{P_1}) = -nRT (1 - \frac{P_2}{P_1})$.
$w = -5 \times 8.314 \times 300 \times (1 - \frac{1}{10}) \, J$.
$w = -12471 \times 0.9 \, J = -11223.9 \, J = -11.224 \, kJ$.

(C) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by: $\Delta U = q + w$.
Given that the internal energy decreases by the same amount as the work done by the system,we have $\Delta U = -w$.
Substituting this into the first law equation: $-w = q + w$.
This implies $q = 0$.
$A$ process in which there is no exchange of heat $(q = 0)$ between the system and the surroundings is known as an adiabatic process.

(D) The reaction is $A_{(g)} \to A_{(l)}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta H = -3RT$.
Here,$\Delta n_g = n_{g, \text{products}} - n_{g, \text{reactants}} = 0 - 1 = -1$.
Substituting the values: $\Delta H = \Delta U - RT$.
$-3RT = \Delta U - RT$.
$\Delta U = -3RT + RT = -2RT$.
Comparing the magnitudes: $|\Delta H| = |-3RT| = 3RT$ and $|\Delta U| = |-2RT| = 2RT$.
Therefore,$|\Delta H| > |\Delta U|$.

(D) For the process $A \rightarrow B$:
Heat absorbed,$q_{AB} = +5\, J$
Work done by the gas,$w_{AB} = -8\, J$ (since work is done by the gas,it is negative in the system's perspective for $\Delta U = q + w$)
Change in internal energy,$\Delta U_{AB} = q_{AB} + w_{AB} = 5 + (-8) = -3\, J$
For the reverse process $B \rightarrow A$:
Since internal energy is a state function,$\Delta U_{BA} = -\Delta U_{AB} = -(-3) = +3\, J$
Heat evolved,$q_{BA} = -3\, J$
Using the first law of thermodynamics: $\Delta U_{BA} = q_{BA} + w_{BA}$
$3 = -3 + w_{BA}$
$w_{BA} = +6\, J$
Since the work done $w_{BA}$ is positive,it means work is done by the surrounding on the gas.

(NONE) The first law of thermodynamics is given by $\Delta U = q + w$.
$1$. For an isochoric process,volume is constant,so $w = 0$,which implies $\Delta U = q$. This is correct.
$2$. For an adiabatic process,$q = 0$,which implies $\Delta U = w$. This is correct.
$3$. For a cyclic process,the internal energy change $\Delta U = 0$,which implies $q = -w$. This is correct.
$4$. For an isothermal process of an ideal gas,$\Delta U = 0$ (since internal energy depends only on temperature),which implies $q = -w$. This is also correct.
Note: All given options are actually correct representations of the first law of thermodynamics for the specified processes. However,if the question implies a standard convention where $w = -P_{ext}\Delta V$,then all statements are valid.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Work is done on the system during compression,so $w = +10 \ kJ$.
Heat is released to the surroundings,so $q = -2 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 10 \ kJ = +8 \ kJ$.

(A) According to the first law of thermodynamics,$\Delta U = Q + W$.
Since the process occurs at constant volume,the work done $W = -P \Delta V = 0$.
Therefore,the change in internal energy $\Delta U$ is equal to the heat absorbed $Q$.
Given $Q = 200 \ J$,the change in internal energy $\Delta U = 200 \ J$.

(C) Both heat $(q)$ and work $(w)$ are path functions,meaning their values depend on the path taken to reach a specific state.
According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by $\Delta U = q + w$.
Since internal energy is a state function,its value depends only on the initial and final states,not on the path taken.
Therefore,$q + w$ represents a state function.

(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since $q = 0$,$\Delta U = w$.
Given $w = -3\, kJ = -3000\, J$ (work done by the system is negative).
We know $\Delta U = n \times C_v \times \Delta T$.
Substituting the values: $-3000 = 1 \times 20 \times (T_f - 300)$.
$-150 = T_f - 300$.
$T_f = 300 - 150 = 150\, K$.

(C) The work done against constant external pressure is given by the formula $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 1 \, atm$,$V_1 = 2 \, L$,$V_2 = 6 \, L$,and $q = +800 \, J$ (heat absorbed).
First,calculate the change in volume: $\Delta V = V_2 - V_1 = 6 \, L - 2 \, L = 4 \, L$.
Convert the work done from $L \cdot atm$ to $J$ using the conversion factor $1 \, L \cdot atm = 101.325 \, J$:
$W = -1 \, atm \times 4 \, L = -4 \, L \cdot atm = -4 \times 101.325 \, J = -405.3 \, J$.
According to the first law of thermodynamics: $\Delta U = q + W$.
Substituting the values: $\Delta U = 800 \, J + (-405.3 \, J) = 394.7 \, J$.
Rounding to the nearest integer,we get $\Delta U \approx 395 \, J$.

(A) For an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.
For an ideal gas,the change in internal energy is $\Delta U = nC_v \Delta T$.
For a monoatomic gas,$C_v = \frac{3}{2}R$.
Given: $n = 0.10 \ mol$,$w = -75 \ cal$ (work done by the system),$T_i = 227 + 273 = 500 \ K$,and $R = 2 \ cal \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values: $n \times \frac{3}{2} \times R \times (T_f - T_i) = w$.
$0.10 \times \frac{3}{2} \times 2 \times (T_f - 500) = -75$.
$0.3 \times (T_f - 500) = -75$.
$T_f - 500 = \frac{-75}{0.3} = -250$.
$T_f = 500 - 250 = 250 \ K$.

(D) For an adiabatic process,the heat exchange $Q = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Since $Q = 0$,$\Delta U = W$.
The work done against a constant external pressure is given by $W = -P_{ext} \Delta V$.
Given $P_{ext} = 0.75 \, atm$,$V_1 = 1 \, L$,and $V_2 = 12 \, L$.
$\Delta V = V_2 - V_1 = 12 \, L - 1 \, L = 11 \, L$.
$W = -0.75 \, atm \times 11 \, L = -8.25 \, atm \cdot L$.
Converting $atm \cdot L$ to Joules $(1 \, atm \cdot L = 101.325 \, J)$:
$\Delta U = -8.25 \times 101.325 \, J = -835.93 \, J$.
Thus,the change in internal energy is approximately $-835.9 \, J$.

(B) According to the first law of thermodynamics,$\Delta U = q + w$.
Given:
Heat absorbed,$q = +84\, kJ$.
Work done by the system,$w = -P \cdot \Delta V$.
Pressure,$P = 1\, MPa = 10^6\, Pa$.
Change in volume,$\Delta V = 0.06\, m^3 - 0.03\, m^3 = 0.03\, m^3$.
Calculating work done:
$w = - (10^6\, Pa) \times (0.03\, m^3) = -30,000\, J = -30\, kJ$.
Calculating change in internal energy:
$\Delta U = 84\, kJ + (-30\, kJ) = 54\, kJ$.

(C) Given: $q = +100 \ J$ (Heat absorbed).
$P_{ext} = 1.50 \ atm$.
$V_1 = 8 \ L, V_2 = 2 \ L$.
$\Delta V = V_2 - V_1 = 2 - 8 = -6 \ L$.
Work done $(w)$ $= -P_{ext} \Delta V = -(1.50 \ atm) \times (-6 \ L) = 9.0 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.325 \ J$,
$w = 9.0 \times 101.325 \ J = 911.925 \ J$.
According to the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = 100 \ J + 911.925 \ J = 1011.925 \ J \approx 1011 \ J$.

(B) In a cyclic process,the system returns to its initial state.
Since state functions depend only on the initial and final states,the change in any state function (such as $\Delta G$,$\Delta S$,or $\Delta H$) is equal to zero.
Work $(w)$ is a path function,not a state function,and its value depends on the path taken.
Therefore,in a cyclic process,the net work done is not necessarily equal to zero.

(C) When ice melts into water at $273 \ K$,the density of water is greater than that of ice,which means the volume of the system decreases $(dV < 0)$.
Since work done $W = P \Delta V$,and $\Delta V$ is negative,the work done by the system on the atmosphere is negative (or work is done on the system by the atmosphere).
Therefore,option $(b)$ is incorrect.
During the phase change,heat is absorbed by the system from the surroundings,which increases the internal energy of the system.
Thus,only statement $(c)$ is correct.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,the heat supplied to the system is $q = +60 \ J$.
The work done by the system is $w = -15 \ J$.
Therefore,$\Delta U = 60 \ J + (-15 \ J) = +45 \ J$.

(A) Given: $q = +1440 \ cal$ (heat absorbed by the system).
Pressure $P = 1 \ atm$.
Change in volume $\Delta V = V_{water} - V_{ice} = 0.0180 \ L - 0.0196 \ L = -0.0016 \ L$.
Work done $w = -P \Delta V = -1 \ atm \times (-0.0016 \ L) = +0.0016 \ L \cdot atm$.
Convert $w$ to calories: $1 \ L \cdot atm = 24.22 \ cal$.
$w = 0.0016 \times 24.22 \ cal \approx 0.0387 \ cal$.
Using the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = 1440 \ cal + 0.0387 \ cal \approx 1440 \ cal$ (since the work done is negligible compared to the heat absorbed).

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: Heat supplied $q = +300 \ J$.
Pressure $P = 1 \ atm$.
Initial volume $V_1 = 1.5 \ L$,Final volume $V_2 = 2 \ L$.
Change in volume $\Delta V = V_2 - V_1 = 2 - 1.5 = 0.5 \ L$.
Work done $w = -P \Delta V = -1 \ atm \times 0.5 \ L = -0.5 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101 \ J$,then $w = -0.5 \times 101 \ J = -50.5 \ J$.
Now,$\Delta U = 300 \ J - 50.5 \ J = 249.5 \ J$.

(B) For an isothermal expansion of an ideal gas in a vacuum (free expansion),the external pressure $P_{ext} = 0$.
Since work done $W = -P_{ext} \Delta V$,the work done is $0$.
According to the first law of thermodynamics,$\Delta U = q + W$.
For an isothermal process of an ideal gas,the internal energy $\Delta U$ depends only on temperature.
Since the temperature is constant $(\Delta T = 0)$,the change in internal energy $\Delta U = 0$.

(C) The balanced chemical equation for the combustion of carbon monoxide is:
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)}$
The change in the number of gaseous moles is given by:
$\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)}$
$\Delta n_g = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$
We know the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is:
$\Delta H = \Delta U + \Delta n_g RT$
Since $\Delta n_g = -0.5$ (which is negative),the term $\Delta n_g RT$ will be negative.
Therefore,$\Delta H = \Delta U - 0.5 RT$.
This implies that $\Delta H < \Delta U$.

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta H$ represents the heat of combustion at constant pressure $(Q_p)$ and $\Delta U$ represents the heat of combustion at constant volume $(Q_v)$.
Substituting these into the equation: $Q_p = Q_v + \Delta n_g RT$.
Given that $\Delta n_g = -3.5$,the equation becomes: $Q_p = Q_v - 3.5RT$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
For a process to have $q = \Delta U$,the work done $(w)$ must be zero.
Work done is given by $w = -P \cdot \Delta V$.
For $w = 0$,the change in volume $(\Delta V)$ must be zero,which corresponds to an isochoric process.
In the given $P-V$ graph,the process $O \to D$ represents a vertical line where the volume $(V)$ remains constant.
Therefore,for the process $O \to D$,$\Delta V = 0$,which implies $w = 0$ and $q = \Delta U$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Given that the work done by the system is equal to the change in internal energy,we have $\Delta U = w$.
Substituting this into the first law equation: $\Delta U = q + \Delta U$,which implies $q = 0$.
$A$ process where heat exchange $q$ is zero is known as an adiabatic process.

(D) In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$,$\Delta U = w$.
In an adiabatic expansion,the system 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 a function of temperature $(U \propto T)$,a decrease in internal energy leads to a decrease in the temperature of the system.

(B) Given: $\Delta H = -92.2\, kJ$,$P = 40\, atm$,$\Delta V = -1\, L$.
Using the relation: $\Delta H = \Delta U + P\Delta V$
Therefore: $\Delta U = \Delta H - P\Delta V$
First,convert $P\Delta V$ from $atm\, L$ to $kJ$ using the conversion factor $1\, atm\, L = 101.325\, J = 0.101325\, kJ$.
$P\Delta V = 40\, atm \times (-1\, L) = -40\, atm\, L$
$P\Delta V = -40 \times 0.101325\, kJ = -4.053\, kJ$
Now,calculate $\Delta U$:
$\Delta U = -92.2\, kJ - (-4.053\, kJ)$
$\Delta U = -92.2\, kJ + 4.053\, kJ = -88.147\, kJ$
Rounding to the nearest integer,we get $\Delta U \approx -88\, kJ$.

(B) For an adiabatic process,the change in internal energy $\Delta U$ is equal to the work done $w$ on the system. According to the first law of thermodynamics,$\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.
Given that the gas expands,work is done by the system,so $w = -3000\, J$.
Using the relation $\Delta U = nC_V\Delta T$,where $n = 1\, mol$ and $C_V = 20\, J/(mol\cdot K)$:
$-3000\, J = 1\, mol \times 20\, J/(mol\cdot K) \times (T_f - T_i)$.
Initial temperature $T_i = 27 + 273 = 300\, K$.
$-3000 = 20 \times (T_f - 300)$.
$-150 = T_f - 300$.
$T_f = 300 - 150 = 150\, K$.

(C) For an isothermal process,the temperature remains constant. Since internal energy $(\Delta U)$ of an ideal gas is a function of temperature only,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Substituting $\Delta U = 0$,we get $0 = Q + W$,which implies $Q = -W$.
This means the heat absorbed by the system is equal to the work done by the system.
However,for an isothermal process,enthalpy change $\Delta H = \Delta U + \Delta(PV)$. For an ideal gas,$\Delta H = \Delta U + \Delta(nRT) = 0 + nR\Delta T = 0$. While $\Delta H$ is zero for an ideal gas,the reason provided is not the fundamental explanation for $Q = -W$,which is based on the first law and $\Delta U = 0$. In general,$\Delta H$ is not necessarily zero for all isothermal processes (e.g.,real gases). Thus,the Assertion is correct,but the Reason is incorrect.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = (1 + 3) - 2 = 2$.
Since $\Delta n_g = 2$ (which is positive),$\Delta H = \Delta E + 2RT$,implying $\Delta H > \Delta E$. Thus,the Assertion is correct.
The Reason states that enthalpy change is always greater than internal energy change,which is false because if $\Delta n_g$ is negative or zero,$\Delta H$ can be less than or equal to $\Delta E$.

(N/A) The first law of thermodynamics states that $\Delta U = q + w$.
$(i)$ Since no heat is absorbed,$q = 0$. Therefore,$\Delta U = w$. The system has an adiabatic wall.
$(ii)$ Since no work is done,$w = 0$. Heat is taken out,so $q$ is negative. Therefore,$\Delta U = -q$. The system has thermally conducting walls (diathermal walls).
$(iii)$ Work is done by the system,so $w$ is negative. Heat is supplied to the system,so $q$ is positive. Therefore,$\Delta U = q - w$. This describes a closed system.

(A) For an expansion into a vacuum,the external pressure $p_{ex} = 0 \ atm$.
The work done in an expansion is given by $w = -p_{ex} \Delta V$.
Since $p_{ex} = 0$,the work done $w = 0 \ L \ atm$.
According to the first law of thermodynamics,$\Delta U = q + w$.
For an isothermal expansion of an ideal gas,the change in internal energy $\Delta U = 0$.
Therefore,$0 = q + 0$,which implies $q = 0$.
Thus,no work is done and no heat is absorbed.

(N/A) The work done during expansion against a constant external pressure is given by the formula $w = -p_{ex} \Delta V$.
Assuming the change in volume $\Delta V$ is $8 \ L$,the work done is $w = -1 \ atm \times 8 \ L = -8 \ L \ atm$.
Since $q = -w$ for an adiabatic process or in the context of energy balance where heat absorbed equals work done,we have $q = 8 \ L \ atm$.

(B) The combustion reaction of methane is: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l).$
The change in the number of gaseous moles is $\Delta n_g = n_p(g) - n_r(g) = 1 - (1 + 2) = -2.$
The relationship between enthalpy change and internal energy change is $\Delta H^{\theta} = \Delta U^{\theta} + \Delta n_g RT.$
Substituting $\Delta n_g = -2,$ we get $\Delta H^{\theta} = \Delta U^{\theta} - 2RT.$
Since $2RT$ is a positive value,$\Delta H^{\theta} < \Delta U^{\theta}.$
Therefore,option $(B)$ is correct.

(A) According to the first law of thermodynamics,$\Delta U = q + W$ $(i)$.
Where,$\Delta U$ is the change in internal energy,$q$ is the heat,and $W$ is the work done.
Given:
$q = +701 \, J$ (heat is absorbed by the system).
$W = -394 \, J$ (work is done by the system).
Substituting these values into equation $(i)$:
$\Delta U = 701 \, J + (-394 \, J) = 307 \, J$.
Therefore,the change in internal energy is $307 \, J$.