First law of thermodynamics Questions in English

(C) The combustion reaction of sucrose is: $C_{12}H_{22}O_{11}(s) + 12O_2(g) \rightarrow 12CO_2(g) + 11H_2O(l)$.
Calculate the change in the number of moles of gaseous species,$\Delta n_g = n_p(g) - n_r(g)$.
Here,$n_p(g) = 12$ (from $12CO_2$) and $n_r(g) = 12$ (from $12O_2$).
Therefore,$\Delta n_g = 12 - 12 = 0$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Since $\Delta n_g = 0$,we get $\Delta H = \Delta U + 0 \times RT$,which implies $\Delta H = \Delta U$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$.
Therefore,$\Delta H - \Delta U = \Delta n_g RT$.
For the given reaction: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$.
The number of moles of gaseous products is $3$ $(CO_2)$.
The number of moles of gaseous reactants is $1 + 5 = 6$ ($C_3H_8$ and $O_2$).
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 3 - 6 = -3$.
Substituting the value of $\Delta n_g$ into the equation: $\Delta H - \Delta U = -3RT$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,the work is done by the system,so $w = -8 \, J$.
The heat is supplied to the system,so $q = +40 \, J$.
Therefore,$\Delta U = 40 \, J + (-8 \, J) = 32 \, J$.

(B) According to the first law of thermodynamics,$\Delta U = q + w$.
For an isothermal process of an ideal gas,$\Delta U = 0$,so $q = -w$.
For a cyclic process,the change in internal energy $\Delta U = 0$,so $q = -w$.
For an adiabatic process,$q = 0$,therefore $\Delta U = w$.
Thus,the relation $q = -w$ is not true for an adiabatic process.

(A) Given: $q = +1.0 \, \text{kcal} = 1000 \, \text{cal}$ (heat added to the system).
Pressure $P = 1 \, \text{atm}$.
Change in volume $\Delta V = V_f - V_i = 1.5 \, \text{L} - 1.2 \, \text{L} = 0.3 \, \text{L}$.
Work done $W = -P \Delta V = -1 \, \text{atm} \times 0.3 \, \text{L} = -0.3 \, \text{L} \cdot \text{atm}$.
Conversion factor: $1 \, \text{L} \cdot \text{atm} = 24.22 \, \text{cal}$.
So,$W = -0.3 \times 24.22 \, \text{cal} = -7.266 \, \text{cal} = -0.007266 \, \text{kcal}$.
Using the first law of thermodynamics: $\Delta U = q + W$.
$\Delta U = 1.0 \, \text{kcal} - 0.007266 \, \text{kcal} = 0.992734 \, \text{kcal} \approx 0.99274 \, \text{kcal}$.

(C) According to the first law of thermodynamics,$\Delta U = q + w$.
For a process occurring at constant volume,the work done $(w)$ is given by $w = -P \Delta V$.
Since the volume is constant,$\Delta V = 0$,which implies $w = 0$.
Therefore,the equation simplifies to $\Delta U = q_{(V)}$,where $q_{(V)}$ is the heat exchanged at constant volume.

(B) An isothermal process is defined as a process where the temperature remains constant,i.e.,$\Delta T = 0$.
For an ideal gas,the internal energy $\Delta U$ is a function of temperature only,so $\Delta U = nC_v\Delta T$.
Since $\Delta T = 0$,it follows that $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $\Delta U = 0$,we have $q = -w$.
Therefore,in an isothermal process,$q \neq 0$ (unless $w = 0$) and $\Delta U = 0$.

(C) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by: $\Delta U = q + w$.
Here,the system absorbs heat,so $q = +450 \ J$.
The system performs work,so $w = -600 \ J$.
Therefore,$\Delta U = 450 \ J - 600 \ J = -150 \ J$.
Since $\Delta U = U_2 - U_1$,we have $U_2 - U_1 = -150 \ J$.
Thus,the final internal energy $U_2 = U_1 - 150 \ J$.

(D) In an adiabatic process,there is no heat exchange with the surroundings,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$,we have $\Delta U = w$.
During expansion,the gas does work on the surroundings,so $w < 0$.
Consequently,the change in internal energy $\Delta U$ is negative,meaning the internal energy of the gas decreases.
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,causing the gas to cool down.

(D) Given: Heat absorbed $q = +1.5 \, J$.
Expansion volume $\Delta V = 500 \, cm^3 - 200 \, cm^3 = 300 \, cm^3 = 0.3 \, L$.
External pressure $P = \frac{750}{760} \, atm \approx 0.9868 \, atm$.
Work done $W = -P_{ext} \Delta V = -0.9868 \, atm \times 0.3 \, L = -0.29604 \, atm \cdot L$.
Converting work to Joules: $W = -0.29604 \times 101.3 \, J \approx -29.99 \, J$ (using standard conversion $1 \, atm \cdot L = 101.3 \, J$).
Using the First Law of Thermodynamics: $\Delta U = q + W$.
$\Delta U = 1.5 \, J - 29.6 \, J = -28.1 \, J$.

(B) The chemical reaction is: $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0.5 = 0.5 \ mol$.
From the first law of thermodynamics,the relationship between enthalpy change and internal energy change is $\Delta H = \Delta E + \Delta n_g RT$.
Therefore,$\Delta H - \Delta E = \Delta n_g RT$.
Substituting the values: $\Delta H - \Delta E = 0.5 \times 8.314 \times 298 = 1238.78 \ J \ mol^{-1}$.

(A) In a bomb calorimeter,the volume is kept constant.
Since $\Delta V = 0$,the work done $w = -P \Delta V = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $w = 0$,$\Delta U = q$.
For an exothermic reaction,the heat released $q$ is negative $(q < 0)$.
Therefore,$\Delta U < 0$ and $w = 0$.

(C) 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,$\Delta n_g$ is the change in the number of moles of gaseous species: $\Delta n_g = (n_{g, products}) - (n_{g, reactants}) = 2 - 3 = -1$.
The temperature $T = 27\,^oC = 300 \,K$.
The gas constant $R = 8.314 \times 10^{-3} \,kJ\,K^{-1} \,mol^{-1}$.
Substituting the values: $-1366.5 = \Delta U + (-1) \times (8.314 \times 10^{-3}) \times 300$.
$-1366.5 = \Delta U - 2.4942$.
$\Delta U = -1366.5 + 2.4942 = -1364.0058 \,kJ\,mol^{-1} \approx -1364 \,kJ\,mol^{-1}$.

(B) For an adiabatic process,$q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$,we have $\Delta U = w$.
During expansion,the system does work on the surroundings,so $w$ is negative $(w < 0)$.
Therefore,$\Delta U$ is negative,which means the internal energy of the system decreases.

(A) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = q + w$. For a cyclic process,$\Delta U = 0$,which implies $q = -w$ or $w = -q$. Here,the total heat absorbed is $q_{total} = Q_1 + Q_2$. Therefore,the work done by the engine should be $W = -(Q_1 + Q_2)$. The given expression $W = Q_1 + Q_2$ implies $W = -q_{total}$,which is consistent with the first law of thermodynamics for a cyclic process where the system returns to its initial state. Thus,the first law is not violated.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: $q = +100 \ cal = 100 \times 4.184 \ J \approx 418.4 \ J$ (Heat supplied to the system is positive).
Work done by the system is $w = -300 \ J$ (Work done by the system is negative).
Therefore,$\Delta U = 418.4 - 300 = 118.4 \ J$.
Rounding to the nearest integer provided in the options,$\Delta U = 120 \ J$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$,which implies $\Delta H - \Delta U = \Delta n_g RT$.
For the reaction $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$:
The number of moles of gaseous products $(n_p)$ = $3$ (for $CO_2$).
The number of moles of gaseous reactants $(n_r)$ = $1 + 5 = 6$ (for $C_3H_8$ and $O_2$).
$\Delta n_g = n_p - n_r = 3 - 6 = -3$.
Therefore,$\Delta H - \Delta U = -3RT$.

(A) The chemical reaction is: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(g)$
The change in the number of moles of gaseous species is calculated as:
$\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 1) = 1 - 2 = -1$
Using the thermodynamic relation:
$\Delta H = \Delta U + \Delta n_g RT$
Substituting $\Delta n_g = -1$:
$\Delta H = \Delta U - RT$
Since $RT$ is positive,it follows that $\Delta H < \Delta U$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the given reaction: $Ag_2O_{(s)} \rightarrow 2Ag_{(s)} + \frac{1}{2}O_{2(g)}$.
The number of moles of gaseous products is $n_p = \frac{1}{2}$ and the number of moles of gaseous reactants is $n_r = 0$.
Therefore,$\Delta n_g = n_p - n_r = \frac{1}{2} - 0 = 0.5$.
Since $\Delta n_g > 0$,the term $\Delta n_g RT$ is positive.
Thus,$\Delta H = \Delta U + (0.5)RT$,which implies $\Delta H > \Delta U$.

(C) Given: $V_1 = 2 \ L$,$V_2 = 2.5 \ L$,$q = +300 \ J$,$P_{ext} = 1 \ atm$.
According to the first law of thermodynamics: $\Delta U = q + W$.
Work done $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1) = -1 \ atm \times (2.5 \ L - 2 \ L) = -0.5 \ L \cdot atm$.
Conversion: $1 \ L \cdot atm = 101.3 \ J$.
So,$W = -0.5 \times 101.3 \ J = -50.65 \ J$.
Change in internal energy: $\Delta U = 300 \ J + (-50.65 \ J) = 249.35 \ J$.

(C) For the free expansion of an ideal gas into a vacuum,the external pressure $P_{ext} = 0$.
According to the first law of thermodynamics,$q = 0$ (adiabatic) and $w = -P_{ext} \Delta V = 0$.
Therefore,the change in internal energy $\Delta U = q + w = 0$.
Since the internal energy of an ideal gas depends only on temperature,$\Delta U = 0$ implies $\Delta T = 0$.
Enthalpy is defined as $H = U + PV$. For an ideal gas,$H = U + nRT$.
Since $\Delta U = 0$ and $\Delta T = 0$,the change in enthalpy $\Delta H = \Delta U + nR\Delta T = 0 + 0 = 0$.
Thus,the enthalpy remains unaffected.

(B) The given reaction is $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$.
The change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - (1 + 3) = 2 - 4 = -2$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta H = -92.38 \ kJ$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting the values: $-92.38 = \Delta U + (-2) \times (8.314 \times 10^{-3}) \times 298$.
$-92.38 = \Delta U - 4.959$.
$\Delta U = -92.38 + 4.959 = -87.421 \ kJ \approx -87.42 \ kJ$.

(B) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: $q = +300 \ J$ (heat absorbed).
Work done $w = -P_{ext} \Delta V$.
$P_{ext} = 1 \ atm = 1.01325 \times 10^5 \ Pa$.
$\Delta V = V_f - V_i = 2.5 \ L - 2.0 \ L = 0.5 \ L = 0.5 \times 10^{-3} \ m^3$.
$w = -(1.01325 \times 10^5 \ Pa) \times (0.5 \times 10^{-3} \ m^3) = -50.66 \ J$.
$\Delta U = 300 \ J - 50.66 \ J = 249.34 \ J$.
Rounding to the nearest provided option,the value is approximately $249.5 \ J$.

(C) Both $q$ (heat) and $w$ (work) are path functions because their values depend on the path taken to reach a state.
However,according to the first law of thermodynamics,$\Delta U = q + w$.
Since the internal energy change $(\Delta U)$ is a state function,the sum $q + w$ is also a state function.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_{(g)}RT$.
For the reaction $C_{(s)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$,the change in the number of moles of gaseous species is $\Delta n_{(g)} = n_{p(g)} - n_{r(g)} = 1 - 1 = 0$.
Since $\Delta n_{(g)} = 0$,the equation becomes $\Delta H = \Delta U + (0)RT$,which simplifies to $\Delta H = \Delta U$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_{(g)}RT$.
For the reaction $3O_{2(g)} \rightarrow 2O_{3(g)}$,the change in the number of moles of gaseous species $(\Delta n_{(g)})$ is calculated as:
$\Delta n_{(g)} = n_{p(g)} - n_{r(g)} = 2 - 3 = -1$.
Substituting this value into the equation:
$\Delta H = \Delta U + (-1)RT$
$\Delta H = \Delta U - RT$.
Rearranging the terms to find $\Delta U - \Delta H$:
$\Delta U - \Delta H = RT$.

(B) The change in the number of gaseous moles is calculated as $\Delta n_g = n_p - n_r = 1 - (1 + 0.5) = -0.5$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Substituting the value of $\Delta n_g$,we get $\Delta H = \Delta U - 0.5 RT$.
Since $\Delta n_g$ is negative,$\Delta H < \Delta U$.

(C) According to the sign convention for the first law of thermodynamics:
Work done on the system $(w) = +5 \ kJ$
Heat released by the system $(q) = -1 \ kJ$
The change in internal energy $(\Delta U)$ is given by the equation:
$\Delta U = q + w$
$\Delta U = -1 \ kJ + 5 \ kJ = 4 \ kJ$

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ for a chemical reaction is given by the equation:
$\Delta H = \Delta E + \Delta n_{g}RT$
Rearranging this equation,we get:
$\Delta H - \Delta n_{g}RT = \Delta E$
Therefore,the correct option is $(B)$.

(D) The conservation of mass and energy is demonstrated by the modified form of the $1^{st}$ law of thermodynamics,which accounts for the equivalence of mass and energy as described by Einstein's equation,$E = mc^2$.

(B) For an irreversible expansion process,the work done is given by $W = -P_{ext} \Delta V$.
Given $P_{ext} = 2.5 \ atm$,$V_i = 2.50 \ L$,and $V_f = 4.50 \ L$.
$W = -2.5 \ atm \times (4.50 \ L - 2.50 \ L) = -2.5 \times 2.0 = -5.0 \ L \ atm$.
Since $1 \ L \ atm = 101.3 \ J$,$W = -5.0 \times 101.3 \ J = -505 \ J$.
As the container is well-insulated,the process is adiabatic,so $q = 0$.
According to the First Law of Thermodynamics,$\Delta U = q + W$.
Therefore,$\Delta U = 0 + (-505 \ J) = -505 \ J$.

(A) The process is $H_2O_{(l)} \longrightarrow H_2O_{(g)}$.
The relation between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta H = 40.66\ kJ\ mol^{-1}$,$T = 100 + 273 = 373\ K$,and $\Delta n_g = 1 - 0 = 1$.
Using $R = 8.314 \times 10^{-3}\ kJ\ K^{-1}\ mol^{-1}$,we have:
$40.66 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.
$40.66 = \Delta U + 3.101$.
$\Delta U = 40.66 - 3.101 = 37.559\ kJ\ mol^{-1} \approx 37.56\ kJ\ mol^{-1}$.

(C) For the free expansion of an ideal gas,the process is adiabatic,so $q = 0$.
Since the expansion occurs against zero external pressure $(P_{ext} = 0)$,the work done is $W = -P_{ext} \Delta V = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$. Since $q = 0$ and $W = 0$,$\Delta U = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only,so $\Delta U = 0$ implies $\Delta T = 0$.

(D) The chemical equation for the formation of $CO$ is:
$C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
From the first law of thermodynamics,we know that $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging the equation,we get $\Delta H - \Delta U = \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species:
$\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting the values:
$\Delta H - \Delta U = \frac{1}{2} \times 8.314 \times 298 = 1238.78 \ J \ mol^{-1}$.

(D) The process is: $H_2O(l) \rightarrow H_2O(g)$.
The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.
Given: $\Delta H = 41\, kJ\, mol^{-1} = 41000\, J\, mol^{-1}$,$T = 373\, K$,and $R = 8.3\, J\, mol^{-1}\, K^{-1}$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$,we get $\Delta U = \Delta H - \Delta n_g RT$.
$\Delta U = 41000\, J\, mol^{-1} - (1 \times 8.3\, J\, mol^{-1}\, K^{-1} \times 373\, K)$.
$\Delta U = 41000 - 3095.9 = 37904.1\, J\, mol^{-1}$.
Converting to $kJ\, mol^{-1}$,we get $\Delta U = 37.9041\, kJ\, mol^{-1}$.

(A) For an isothermal reversible expansion of an ideal gas,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 $q = -w$.
Given that the system absorbs $208 \ J$ of heat,$q = +208 \ J$.
Therefore,$w = -q = -208 \ J$.
Thus,the values are $q = +208 \ J$ and $w = -208 \ J$.

(C) According to the first law of thermodynamics,$\Delta U = Q + W$.
In an adiabatic process,there is no exchange of heat between the system and the surroundings,so $Q = 0$.
Therefore,$\Delta U = W_{\text{adiabatic}}$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species.
For the reaction: $A_2B_{(s)} \to 2A_{(s)} + 1/2B_{2(g)}$,the value of $\Delta n_g = (1/2) - 0 = 0.5 \ mol$.
Given: $\Delta H = +7.3 \ kcal$,$T = 298 \ K$,and $R \approx 0.002 \ kcal \ K^{-1} \ mol^{-1}$.
Substituting the values: $\Delta E = \Delta H - \Delta n_g RT = 7.3 - (0.5 \times 0.002 \times 298) = 7.3 - 0.298 = 7.002 \ kcal$.
Since $7.002 \ kcal < 7.3 \ kcal$,the heat change at constant volume $(\Delta E)$ is less than $7.3 \ kcal$.

(B) For an adiabatic process, the work done $(w)$ is equal to the change in internal energy $(\Delta U)$.
$w = \Delta U = n C_V \Delta T$
Given:
$n = 2 \ mol$
$C_V = 12.5 \ J/K/mol$
$\Delta T = T_f - T_i = 200 \ K - 300 \ K = -100 \ K$
$w = 2 \ mol \times 12.5 \ J/K/mol \times (-100 \ K)$
$w = -2500 \ J$
Converting to $kJ$:
$w = -2500 / 1000 = -2.5 \ kJ$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,$q = +300 \ J$ (heat is provided to the system).
The work done $w = -P_{ext} \Delta V$.
Given $P_{ext} = 1 \ atm = 101.3 \ kPa$ and $\Delta V = (2 - 1.5) \ L = 0.5 \ L$.
$w = -1 \ atm \times 0.5 \ L = -0.5 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.3 \ J$,$w = -0.5 \times 101.3 \ J = -50.65 \ J$.
Therefore,$\Delta U = 300 \ J - 50.65 \ J = 249.35 \ J$.
Rounding to the nearest provided option,the value is $249.5 \ J$.

(B) For a cyclic process,the change in internal energy $\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = Q_{net} + W_{net} = 0$,which implies $Q_{net} = -W_{net}$.
$Q_{net} = Q_1 + Q_2 + Q_3 + Q_4 = 6000 - 5500 - 3000 + 3500 = 1000 \, J$.
$W_{net} = W_1 + W_2 + W_3 + W_4 = 2500 - 1000 - 1200 + x = 300 + x$.
Since $Q_{net} = -W_{net}$,we have $1000 = -(300 + x)$,which gives $x = -1300 \, J$. Thus,$|x| = 1300 \, J$.
The total heat absorbed by the gas is the sum of positive heat values: $Q_{abs} = Q_1 + Q_4 = 6000 + 3500 = 9500 \, J$.
The net work done is $W_{net} = 300 + (-1300) = -1000 \, J$. The magnitude of net work done is $|W_{net}| = 1000 \, J$.
$\eta = \frac{|W_{net}|}{Q_{abs}} = \frac{1000}{9500} \approx 0.10526 = 10.53\%$.

(B) For an ideal gas reaction,the relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by $\Delta H = \Delta U + \Delta n_g RT$.
In the reaction $A_{2(g)} + B_{2(g)} \rightarrow 2AB_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,the equation becomes $\Delta H = \Delta U + (0)RT$,which implies $\Delta H = \Delta U$.
Therefore,option $B$ is incorrect as it states $\Delta H \neq \Delta U$.
Option $A$ and $C$ are not necessarily true for all such reactions.
According to Kirchhoff's law,$\left[ \frac{\partial (\Delta_r H)}{\partial T} \right]_P = \Delta C_p$.
For this specific reaction,$\Delta C_p = C_p(2AB) - [C_p(A_2) + C_p(B_2)]$.
While $\Delta C_p$ is generally not zero,in the context of standard textbook problems regarding ideal gas reactions where $\Delta n_g = 0$,the relationship $\Delta H = \Delta U$ is the fundamental identity. However,looking at the options provided,if $\Delta n_g = 0$,then $\Delta H = \Delta U$ is the correct identity. Since the question asks what 'must' be correct and option $B$ says they are not equal,$B$ is false. Given the constraints,the question likely implies identifying the relationship between $\Delta H$ and $\Delta U$.

(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 both $q = 0$ and $w = 0$,the change in internal energy $\Delta U = 0 + 0 = 0$.
Therefore,the internal energy change for an ideal gas during adiabatic free expansion is zero.

(C) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_{g}RT$.
Given:
$\Delta H = 10 \ kcal/mole = 10000 \ cal/mole$.
$T = 350 \ K$.
$P = 2 \ atm$.
For the phase change $Liquid \rightarrow Vapour$,$\Delta n_{g} = 1$ (assuming ideal gas behavior for vapour).
Using the ideal gas law $PV = nRT$,we have $\Delta n_{g}RT = P\Delta V = \Delta(PV)$.
However,for a phase change,$\Delta H = \Delta U + P\Delta V$.
Assuming the volume of liquid is negligible compared to vapour,$P\Delta V \approx P V_{g} = nRT$.
Substituting the values:
$10000 = \Delta U + (1 \times 1.987 \times 350) \approx \Delta U + 700$.
Using $R \approx 2 \ cal \ K^{-1} \ mol^{-1}$ as per standard convention for such problems:
$10000 = \Delta U + (1 \times 2 \times 350)$.
$10000 = \Delta U + 700$.
$\Delta U = 10000 - 700 = 9300 \ cal$.

(A) For an ideal gas,the internal energy $U$ is a function of temperature $T$ only.
In an adiabatic process,the first law of thermodynamics is given by $\Delta U = q + w$. Since $q = 0$,we have $\Delta U = w$.
In compression,work is done on the system,so $w > 0$.
Therefore,$\Delta U > 0$,which means the internal energy increases during adiabatic compression.

(D) The $1^{st}$ law of thermodynamics is a statement of the law of conservation of energy.
It states that energy can neither be created nor be destroyed,although it can be transformed from one form to another.
For an isolated system,the total energy remains constant.
For a closed system,the change in internal energy is given by $\Delta U = q + w$,which is consistent with the conservation of energy principle.
Therefore,both the statement about the law of conservation of energy and the concept of energy in a closed system (in the context of energy balance) are fundamental aspects of this law.

(D) The container is thermally insulated,so there is no heat exchange with the surroundings,meaning $q = 0$.
Work is done on the system by the mechanical stirrer,so by the sign convention,$w > 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$ and $w > 0$,it follows that $\Delta U = w > 0$.
Therefore,the internal energy of the system increases.

(B) The relationship between enthalpy change $(\Delta H^o)$ and internal energy change $(\Delta U^o)$ is given by the equation: $\Delta H^o = \Delta U^o + \Delta n_g RT$.
First,calculate the change in the number of moles of gaseous species $(\Delta n_g)$:
$\Delta n_g = \sum n_{g, \text{products}} - \sum n_{g, \text{reactants}}$
$\Delta n_g = 2 - (1 + 1) = 2 - 2 = 0$.
Since $\Delta n_g = 0$,the equation becomes:
$\Delta H^o = \Delta U^o + (0)RT$
$\Delta H^o = \Delta U^o$.
Given $\Delta H^o = -185 \ kJ$,therefore $\Delta U^o = -185 \ kJ$.

(C) The heat absorbed by the system is $Q = +80 \ kJ$.
The work done by the gas during expansion is given by $W = -P_{ext} \Delta V$.
Given $P = 25 \ bar$ and $\Delta V = (10 \ L - 2 \ L) = 8 \ L$.
$W = -25 \ bar \times 8 \ L = -200 \ bar-L$.
Since $1 \ bar-L = 100 \ J = 0.1 \ kJ$,we have $W = -200 \times 0.1 \ kJ = -20 \ kJ$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
$\Delta U = 80 \ kJ + (-20 \ kJ) = 60 \ kJ$.