First law of thermodynamics Questions in English

(D) In an adiabatic process,there is no exchange of heat between the system and the surroundings $(q = 0)$.
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 $(w < 0)$,which leads to a decrease in internal energy $(\Delta U < 0)$.
For an ideal gas,internal energy is a function of temperature only $(U = f(T))$,so a decrease in internal energy results in a decrease in temperature.
This is also related to the Joule-Thomson effect,where a gas expanding adiabatically from a high-pressure region to a low-pressure region experiences a temperature drop.

(D) The correct statement is that the total energy of an isolated system is constant,which is the definition of the First Law of Thermodynamics.
For a reaction to be feasible,$\Delta G$ must be negative (spontaneous).
Entropy is a measure of disorder,not order.
Therefore,option $D$ is correct.

(A) For an isochoric process,the change in volume is $\Delta V = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$,where $w = -P_{ext} \Delta V$.
Since $\Delta V = 0$,the work done $w = 0$.
Therefore,$\Delta U = q_v$,which means the increase in internal energy is equal to the heat absorbed at constant volume.

(D) The first law of thermodynamics is a statement of the law of conservation of energy,which states that energy can neither be created nor destroyed,only transformed from one form to another. In the context of nuclear reactions or mass-energy equivalence $(E = mc^2)$,it is also associated with the conservation of mass-energy. Therefore,option $(d)$ is the most appropriate choice in this context.

(B) The chemical equation for the reaction is: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$.
Calculate the change in the number of moles of gaseous species,$\Delta n_g = n_p - n_r = 2 - (2 + 1) = 2 - 3 = -1$.
The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by: $\Delta H = \Delta E + \Delta n_g RT$.
Substituting $\Delta n_g = -1$,we get: $\Delta H = \Delta E - RT$.
Since $R$ and $T$ are positive values,$\Delta H = \Delta E - RT$ implies that $\Delta H < \Delta E$.

(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 $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is calculated as: $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$.
Since $\Delta n_g = -0.5$ (which is negative),the term $\Delta n_g RT$ is negative.
Therefore,$\Delta H = \Delta E - 0.5 RT$,which implies that $\Delta H < \Delta E$.

(D) The relationship between enthalpy change and internal energy change for a gaseous reaction is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous products and reactants,defined as $\Delta n_g = n_p - n_r$.
If the number of moles of products is less than the number of moles of reactants,then $\Delta n_g$ is negative $(\Delta n_g < 0)$.
In this case,$\Delta H = \Delta E - |\Delta n_g| RT$,which implies $\Delta H < \Delta E$.
Therefore,the correct option is $D$.

(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 products and reactants.
For the reaction $N_2(g) + 3H_2(g) \to 2NH_3(g)$,the number of moles of gaseous products is $n_p = 2$ and the number of moles of gaseous reactants is $n_r = 1 + 3 = 4$.
Therefore,$\Delta n_g = n_p - n_r = 2 - 4 = -2$.
Substituting this into the equation,we get $\Delta H = \Delta U - 2RT$.
Since $R$ and $T$ are positive values,it follows that $\Delta H < \Delta U$.

(A) The first law of thermodynamics states that the change in internal energy $(\Delta U)$ of a system is equal to the sum of heat added to the system $(q)$ and the work done on the system $(w)$.
Mathematically,this is expressed as $\Delta U = q + w$.

(B) For a reversible isothermal expansion,the work done is given by $W = 2.303 \, nRT \log \frac{V_2}{V_1}$.
Given: $n = 1 \, \text{mole}$,$T = 25 + 273 = 298 \, K$,$V_1 = 10 \, L$,$V_2 = 20 \, L$.
Since $R = 8.314 \, J \, K^{-1} \, mol^{-1}$ and $1 \, J = 10^7 \, \text{ergs}$,$R = 8.314 \times 10^7 \, \text{ergs} \, K^{-1} \, mol^{-1}$.
Substituting the values: $W = 2.303 \times 1 \times (8.314 \times 10^7) \times 298 \times log \frac{20}{10}$.
$W = 298 \times 10^7 \times 8.314 \times 2.303 \log \, 2 \, \text{ergs}$.

(C) The law of conservation of energy,also known as the $First \ Law \ of \ Thermodynamics$,states that energy can neither be created nor destroyed,although it can be transformed from one form to another.

(C) The first law of thermodynamics states that the change in internal energy ($\Delta E$ or $\Delta U$) of a system is equal to the sum of heat added to the system $(q)$ and the work done on the system $(W)$.
Therefore,the mathematical expression is $\Delta E = q + W$.

(B) The Joule-Thomson expansion is a process in which a gas is allowed to expand through a porous plug or a throttle into a region of lower pressure.
Since the process is adiabatic $(q = 0)$ and no external work is done $(w = 0)$,the change in internal energy is zero $(\Delta U = 0)$.
From the first law of thermodynamics,$\Delta H = \Delta U + \Delta (PV)$.
For this process,it is observed that the enthalpy remains constant.
Therefore,the Joule-Thomson expansion is an isoenthalpic process,meaning $\Delta H = 0$.

(B) According to the first law of thermodynamics,$\Delta E = q + W$.
For an adiabatic process,the heat exchange $q = 0$.
Substituting $q = 0$ into the equation,we get $\Delta E = W$.

(A) 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 $CH_3COOH_{(l)} + 2O_{2(g)} \rightleftharpoons 2CO_{2(g)} + 2H_2O_{(l)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{g(products)} - n_{g(reactants)} = 2 - 2 = 0$.
Since $\Delta n_g = 0$,the equation becomes $\Delta H = \Delta E$.
Therefore,$\Delta E = \Delta H = -874\,kJ$.

(C) According to the first law of thermodynamics,$\Delta E = Q + W$.
Since the container is insulated,there is no exchange of heat with the surroundings,so $Q = 0$.
Work is done on the system by the paddle,so $W \neq 0$.
Therefore,the change in internal energy is equal to the work done,$\Delta E = W \neq 0$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which implies $\Delta H - \Delta E = \Delta n_g RT$.
For the reaction $C_3H_{8(g)} + 5O_{2(g)} \to 3CO_{2(g)} + 4H_2O_{(l)}$,the change in the number of gaseous moles $(\Delta n_g)$ is calculated as: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.
$\Delta n_g = 3 - (1 + 5) = 3 - 6 = -3$.
Substituting this value into the equation: $\Delta H - \Delta E = -3RT$.

(C) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: Heat supplied to the system,$q = +40 \ J$.
Work done by the system,$w = -8 \ J$ (since work is done by the system,it is negative).
Therefore,the change in internal energy is $\Delta U = 40 \ J + (-8 \ J) = 32 \ J$.

(A) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,the system absorbs heat,so $q = +600 \ J$.
The system does work on the surroundings,so $w = -300 \ J$.
Substituting these values into the equation: $\Delta U = 600 \ J + (-300 \ J) = 300 \ J$.
Therefore,the change in internal energy is $300 \ J$.

(B) 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 $PCl_{5(g)} \to PCl_{3(g)} + Cl_{2(g)}$,the change in the number of gaseous moles $(\Delta n_g)$ is calculated as: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = (1 + 1) - 1 = 1$.
Since $\Delta n_g = 1$,which is greater than $0$,the term $\Delta n_g RT$ is positive.
Therefore,$\Delta H = \Delta E + RT$,which implies $\Delta H > \Delta E$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ for a gaseous reaction is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $N_2O_{4(g)} \to 2NO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - 1 = 1$.
Since $\Delta n_g = 1$ (which is positive),the term $\Delta n_g RT$ is positive.
Therefore,$\Delta H = \Delta E + RT$,which implies $\Delta H > \Delta E$.

(B) 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 $\Delta H = \Delta E$ to hold true,the change in the number of moles of gaseous products and reactants must be zero,i.e.,$\Delta n_g = 0$.
In option $B$,$N_2(g) + O_2(g) \to 2NO(g)$,the number of moles of gaseous products is $2$ and the number of moles of gaseous reactants is $1 + 1 = 2$.
Thus,$\Delta n_g = 2 - 2 = 0$.
Therefore,$\Delta H = \Delta E + (0)RT = \Delta E$.

(A) For an adiabatic process,the first law of thermodynamics states that $\Delta U = q + w$. Since the process is adiabatic,$q = 0$,so $\Delta U = w$.
For an ideal gas,$\Delta U = n C_v \Delta T$. For a monoatomic gas,$C_v = \frac{3}{2} R$.
Thus,$n \times \frac{3}{2} R \times (T_f - T) = -P_{ext} \times \Delta V$.
Given $n = 1 \, \text{mole}$,$P_{ext} = 1 \, \text{atm}$,$\Delta V = (2 - 1) \, \text{L} = 1 \, \text{L}$,and $R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}$.
Substituting the values: $1 \times \frac{3}{2} \times 0.0821 \times (T_f - T) = -1 \times 1$.
$(T_f - T) = -\frac{2}{3 \times 0.0821}$.
$T_f = T - \frac{2}{3 \times 0.0821}$.

(D) The $1^{st}$ law of thermodynamics states that the change in internal energy $\Delta U = q + w$. For a cyclic process,$\Delta U = 0$,so $q = -w$. Here,the total heat absorbed is $Q_1 + Q_2$ and the work done is $W = Q_1 + Q_2$. Since $W = Q_{total}$,the $1^{st}$ law is satisfied. However,this process violates the $2^{nd}$ law of thermodynamics,which states that heat cannot be completely converted into work in a cyclic process without any other effect.

(C) According to the first law of thermodynamics,$\Delta U = q + W$.
Since work is done on the system (compression),$W = + 462 \, J$.
Since energy is evolved (heat released),$q = - 128 \, J$.
Therefore,$\Delta U = (- 128 \, J) + (+ 462 \, J) = + 334 \, J$.

(A) The combustion reaction for butanol is:
$C_4H_9OH_{(l)} + 6O_{2(g)} \rightarrow 4CO_{2(g)} + 5H_2O_{(l)}$
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 given reaction:
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 4 - 6 = -2$
Since $\Delta n_g = -2$ (which is negative),we have:
$\Delta H = \Delta E - 2RT$
Therefore,$\Delta H < \Delta E$.

(D) In an adiabatic process,the system is thermally isolated,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $q = 0$,$\Delta U = w$.
During expansion,the gas does work on the surroundings,so $w < 0$.
This implies $\Delta U < 0$,meaning the internal energy of the gas decreases.
Since the internal energy of an ideal gas is a function of temperature only $(U \propto T)$,a decrease in internal energy leads to a decrease in temperature,causing the gas to cool.

(B) For an isobaric process,the heat supplied is $dQ = n C_p dT$ and the work done is $dW = n R dT$.
The ratio is $\frac{dQ}{dW} = \frac{n C_p dT}{n R dT} = \frac{C_p}{R}$.
For a diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7R}{2}$.
Therefore,$\frac{dQ}{dW} = \frac{7R/2}{R} = \frac{7}{2}$ or $7:2$.

(A) The chemical equation for the formation of $CO$ from its elements is:
$C(s) + \frac{1}{2} O_2(g) \to CO(g)$
We know the relation $\Delta H = \Delta U + \Delta n_g RT$,which implies $\Delta H - \Delta U = \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species:
$\Delta n_g = n_p(g) - n_r(g) = 1 - 0.5 = 0.5$.
Substituting the values:
$\Delta H - \Delta U = 0.5 \times 8.314 \times 298 = 1238.78 \, J \, mol^{-1}$.

(B) The given reaction is: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
The change in the number of gaseous moles is calculated as: $\Delta n_g = n_{products(g)} - n_{reactants(g)} = 2 - (1 + 3) = 2 - 4 = -2$
The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the formula: $\Delta H = \Delta U + \Delta n_g RT$
Substituting the value of $\Delta n_g$: $\Delta H = \Delta U + (-2)RT = \Delta U - 2RT$

(D) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$,the change in the number of moles of gaseous species is calculated as: $\Delta n_g = \sum n_p - \sum n_r = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting $\Delta n_g = -2$ into the equation,we get: $\Delta H = \Delta U - 2RT$.
Since $2RT$ is a positive value,it follows that $\Delta H < \Delta U$.

(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 $C_{(s)} + \frac{1}{2}O_{2(g)} \to CO_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = n_{g(products)} - n_{g(reactants)}$.
Here,$n_{g(products)} = 1$ (for $CO_{(g)}$) and $n_{g(reactants)} = \frac{1}{2}$ (for $O_{2(g)}$).
Therefore,$\Delta n_g = 1 - 0.5 = 0.5$.
Since $\Delta n_g > 0$,the term $\Delta n_g RT$ is positive.
Thus,$\Delta H = \Delta U + 0.5RT$,which implies $\Delta H > \Delta U$.

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For $\Delta H = \Delta U$,the term $\Delta n_g$ must be equal to $0$.
$\Delta n_g$ is the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants.
For option $A$: $H_{2(g)} + Br_{2(g)} \rightarrow 2 HBr_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$.
Since $\Delta n_g = 0$,$\Delta H = \Delta U$ for this reaction.

(A) The first law of thermodynamics states that the change in internal energy $(\Delta U)$ of a system is equal to the heat added to the system $(q)$ minus the work done by the system $(W)$.
Mathematically,this is expressed as $\Delta U = q - W$.

(B) 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 $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is: $\Delta n_{(g)} = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$.
Since $\Delta n_{(g)} < 0$,it follows that $\Delta H < \Delta U$.

(C) Given: $P = 1 \ atm$,$V_1 = 2 \ L$,$V_2 = 6 \ L$,$q = +800 \ J$ (heat absorbed).
Change in volume $\Delta V = 6 - 2 = 4 \ L$.
Work done $W = -P \Delta V = -1 \ atm \times 4 \ L = -4 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.325 \ J$,then $W = -4 \times 101.325 = -405.3 \ J$.
According to the first law of thermodynamics: $\Delta U = q + W$.
$\Delta U = 800 \ J + (-405.3 \ J) = 394.7 \ J$.
Rounding to the nearest provided option,the change in internal energy is $394.95 \ J$.

(D) 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 $\Delta U = q$.
Given that the heat supplied $q = 500 \ J$,the change in internal energy $\Delta U = 500 \ J$.

(B) The given reaction is $A_{(g)} + 3B_{(g)} \rightarrow 3C_{(g)} + 3D_{(g)}$.
The change in the number of moles of gaseous products and reactants is $\Delta n_g = (3 + 3) - (1 + 3) = 6 - 4 = 2$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Given $\Delta U = 17 \, Kcal$,$R = 2 \, cal \, K^{-1} \, mol^{-1} = 2 \times 10^{-3} \, Kcal \, K^{-1} \, mol^{-1}$,and $T = 27 + 273 = 300 \, K$.
Substituting the values: $\Delta H = 17 + (2 \times 2 \times 10^{-3} \times 300) = 17 + 1.2 = 18.2 \, Kcal$.

(A) The chemical equation for the vaporization of water is: $H_2O_{(l)} \to H_2O_{(g)}$
The change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.
The relationship between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_g RT$.
Given: $\Delta H = 40.63 \, kJ \, mol^{-1}$,$R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$,and $T = 100 + 273 = 373 \, K$.
Substituting the values: $40.63 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 373)$.
$40.63 = \Delta U + 3.10$.
$\Delta U = 40.63 - 3.10 = 37.53 \, kJ \, mol^{-1}$.

(C) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta(PV)$.
Given $\Delta H = -560 \, kJ$.
We need to find $\Delta(PV) = P_2V_2 - P_1V_1$.
Initial state: $P_1 = 70 \, atm$,$V_1 = 1 \, L$,so $P_1V_1 = 70 \, atm \cdot L$.
Final state: $P_2 = 40 \, atm$,$V_2 = 1 \, L$,so $P_2V_2 = 40 \, atm \cdot L$.
Thus,$\Delta(PV) = (40 - 70) \, atm \cdot L = -30 \, atm \cdot L$.
Using the conversion factor $1 \, atm \cdot L = 0.1 \, kJ$,we get $\Delta(PV) = -30 \times 0.1 \, kJ = -3 \, kJ$.
Substituting into the equation: $-560 \, kJ = \Delta U + (-3 \, kJ)$.
Therefore,$\Delta U = -560 + 3 = -557 \, kJ$.

(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$,the change in the number of gaseous moles is $\Delta n_g = 2 - (1 + 3) = -2$.
Given: $\Delta H = -92.38 \ kJ$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,and $T = 298 \ K$.
Substituting the values: $-92.38 = \Delta U + (-2) \times (8.314 \times 10^{-3}) \times 298$.
$-92.38 = \Delta U - 4.955$.
$\Delta U = -92.38 + 4.955 = -87.425 \ kJ \approx -87.43 \ kJ$.

(A) At constant volume,the change in volume $\Delta V = 0$.
According to the first law of thermodynamics,$W = -P \Delta V = 0$.
Since $\Delta U = q + W$,and $W = 0$,then $\Delta U = q$.
Given that the heat absorbed $q = 200 \ J$,the change in internal energy $\Delta U = 200 \ J$.

(B) For an adiabatic process,the work done by the system is given by $W = -nC_v(T_2 - T_1)$.
Given: $n = 1 \, mol$,$T_1 = 27^oC = 300 \, K$,$W = 3 \, kJ = 3000 \, J$,and $C_v = 20 \, J/K \cdot mol$.
Since the work is done by the system,$W$ is positive in the formula $W = -\Delta U = -nC_v(T_2 - T_1)$.
Substituting the values: $3000 = -1 \times 20 \times (T_2 - 300)$.
$3000 = -20T_2 + 6000$.
$20T_2 = 3000$.
$T_2 = 150 \, K$.

(B) The chemical reaction is: $2CO(g) + O_2(g) \rightarrow 2CO_2(g)$.
The relationship between enthalpy change and internal energy change is given by: $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous products and reactants: $\Delta n_g = n_p - n_r = 2 - (2 + 1) = 2 - 3 = -1$.
Since $\Delta n_g = -1$,the equation becomes: $\Delta H = \Delta U - RT$.
Therefore,$\Delta H < \Delta U$.

(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 $\Delta H \neq \Delta U$,the condition $\Delta n_g \neq 0$ must be satisfied,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
$(A)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(B)$ $\Delta n_g = 0$ (as all reactants and products are in aqueous or liquid state).
$(C)$ $\Delta n_g = 1 - 1 = 0$.
$(D)$ $\Delta n_g = 2 - (1 + 3) = -2$.
Since $\Delta n_g \neq 0$ only in option $(D)$,the correct answer is $(D)$.

(B) The combustion reaction of isobutane $(C_4H_{10})$ is:
$C_4H_{10(g)} + \frac{13}{2}O_{2(g)} \to 4CO_{2(g)} + 5H_2O_{(g)}$
Calculate the change in the number of moles of gaseous species $(\Delta n_g)$:
$\Delta n_g = (n_{products}) - (n_{reactants}) = (4 + 5) - (1 + 6.5) = 9 - 7.5 = 1.5$
Since $\Delta n_g > 0$,using the relation $\Delta H = \Delta U + \Delta n_g RT$,we get $\Delta H > \Delta U$.

(B) The chemical equation is $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$.
The change in the number of moles of gaseous species is $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0 = 1$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Given $T = 977 + 273 = 1250 \, K$ and $R = 8.314 \times 10^{-3} \, kJ \cdot K^{-1} \cdot mol^{-1}$.
Substituting the values: $174 = \Delta U + (1 \times 8.314 \times 10^{-3} \times 1250)$.
$174 = \Delta U + 10.3925$.
$\Delta U = 174 - 10.3925 = 163.6075 \, kJ \approx 163.6 \, kJ$.

(B) For a process at constant volume,the work done $w = P \Delta V = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since $w = 0$,we have $\Delta U = q$.
Given that the heat supplied $q = 500 \, J$,therefore $\Delta U = 500 \, J$ and $w = 0$.

(A) For an adiabatic process,$q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
Since the gas is compressed,work is done on the system,so $w > 0$.
Therefore,$\Delta U > 0$,which means the internal energy of the system increases.
Since internal energy is a function of temperature $(U \propto T)$,an increase in internal energy leads to an increase in temperature.
Thus,the final temperature is greater than the initial temperature.