First law of thermodynamics Questions in English

(B) The given reaction is: $2 SO_{3(g)} \longrightarrow 2 SO_{2(g)} + O_{2(g)}$
Calculate the change in the number of moles of gaseous species,$\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)}$
$\Delta n_g = (2 + 1) - 2 = 1$
Using the thermodynamic relation: $\Delta H = \Delta U + \Delta n_g RT$
Substituting $\Delta n_g = 1$,we get: $\Delta H = \Delta U + RT$
Therefore,$\Delta H - \Delta U = RT$

(D) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that heat is released from the system,$Q = -2 \ kJ$.
Since work is done on the system,$W = +6 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 6 \ kJ = +4 \ kJ$.

(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 reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,the change in the number of moles of gaseous species $(\Delta n_g)$ is calculated as: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting this value into the equation: $\Delta H = \Delta U + (-2) RT = \Delta U - 2 RT$.
Therefore,the correct option is $D$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species,calculated as: $\Delta n_g = \sum n_{g, \text{products}} - \sum n_{g, \text{reactants}}$.
For the reaction $C_2H_5OH_{(l)} + 3O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_2O_{(l)}$,the gaseous moles are:
Products: $2 \text{ moles of } CO_2$.
Reactants: $3 \text{ moles of } O_2$.
$\Delta n_g = 2 - 3 = -1$.
Substituting this value into the equation,we get: $\Delta H = \Delta E + (-1)RT = \Delta E - RT$.

(D) Statement-$I$: In an adiabatic process,$q = 0$,so according to the first law of thermodynamics,$\Delta U = w$. When work is done on the system,$w$ is positive,which leads to an increase in internal energy $(\Delta U > 0)$. Thus,Statement-$I$ is true.
Statement-$II$: Free expansion occurs against zero external pressure $(P_{ext} = 0)$. Since $w = -P_{ext} \Delta V$,the work done is zero. Thus,Statement-$II$ is true.

(D) The first law of thermodynamics is given by the equation: $\Delta U = Q + W$.
For a cyclic process,the system returns to its initial state,so the change in internal energy is zero: $\Delta U = 0$.
Substituting this into the first law equation: $0 = Q + W$,which simplifies to $Q = -W$.
Therefore,the correct statement is that for a cyclic process,$Q = -W$.

(A) The first law of thermodynamics is given by $\Delta E = q + w$.
$(A)$ For an isochoric process,$\Delta V = 0$,so the work done $w = -P_{ext} \Delta V = 0$. Therefore,$\Delta E = q$.
$(B)$ For an adiabatic process,$q = 0$,so $\Delta E = w$.
$(C)$ For an isothermal process,$\Delta T = 0$,which implies $\Delta E = 0$ for an ideal gas,so $q = -w$.
$(D)$ For a cyclic process,the change in state function $\Delta E = 0$,therefore $q = -w$.

(D) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
For $\Delta H > \Delta U$,the value of $\Delta n_g$ (change in the number of moles of gaseous products and reactants) must be positive,i.e.,$\Delta n_g > 0$.
In option $(d)$: $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$.
Here,$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 1 - 0 = 1$.
Since $\Delta n_g$ is positive,$\Delta H > \Delta U$.

(C) According to the $1^{st}$ Law of Thermodynamics,energy can neither be created nor destroyed,meaning the total energy of the universe remains constant.
According to the $2^{nd}$ Law of Thermodynamics,the entropy of the universe is continuously increasing for any spontaneous process.
Therefore,the statement that the total energy of the universe remains constant is true.

(C) The chemical reaction is: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(g)$
The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$
Where $\Delta n_g$ is the change in the number of moles of gaseous species.
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta n_g = 1 - (1 + 1) = 1 - 2 = -1$
Substituting this into the equation: $\Delta H = \Delta U - RT$
Since $RT$ is positive,it follows that $\Delta H < \Delta U$.

(C) For an isothermal process of an ideal gas,$\Delta U = 0$.
According to the first law of thermodynamics,$\Delta U = q + w = 0$,which implies $w = -q$.
Given that the system releases $460.6 \ L \ atm$ of heat,$q = -460.6 \ L \ atm$.
Therefore,$w = -(-460.6 \ L \ atm) = +460.6 \ L \ atm$.
For a reversible isothermal compression,the work done is given by $w = -2.303 nRT \log(\frac{V_2}{V_1})$.
Since $PV = nRT$,we can write $w = -2.303 P_1 V_1 \log(\frac{V_2}{V_1})$.
Substituting the given values: $460.6 = -2.303 \times (2 \ atm \times 100 \ L) \log(\frac{X}{100})$.
$460.6 = -460.6 \log(\frac{X}{100})$.
$-1 = \log(\frac{X}{100})$.
$\frac{X}{100} = 10^{-1} = 0.1$.
$X = 100 \times 0.1 = 10 \ L$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For $\Delta H \neq \Delta U$,we must have $\Delta n_g \neq 0$.
$(A)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(B)$ $\Delta n_g = (1 + 1) - 2 = 0$.
$(C)$ $\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$. Since $\Delta n_g \neq 0$,$\Delta H \neq \Delta U$.
$(D)$ $\Delta n_g = 1 - (0 + 1) = 0$ (Note: $C_{(s)}$ is a solid,so its moles are not counted in $\Delta n_g$).
Therefore,the correct option is $C$.

(A) Given: $V_1 = 10.0 \ L$,$V_2 = 2.0 \ L$,$P_{ext} = 2.0 \ atm$,$q = -900 \ J$ (heat evolved).
Work done $(w)$ = $-P_{ext}(V_2 - V_1) = -2.0 \times (2.0 - 10.0) = 16.0 \ L \ atm$.
Converting work to Joules: $w = 16.0 \times 101.3 \ J = 1620.8 \ J$.
According to the first law of thermodynamics: $\Delta U = q + w$.
$\Delta U = -900 + 1620.8 = 720.8 \ J$.

(D) For an isothermal reversible expansion of an ideal gas,the heat absorbed $(q_{rev})$ is equal to the work done by the system $(-w_{rev})$.
$q_{rev} = -w_{rev} = 2.303 P_1 V_1 \log \frac{V_2}{V_1}$
Given $P_1 = 20 \ atm$,$V_1 = 1 \ L$,and $q_{rev} = 92.12 \ L \ atm$.
$92.12 = 2.303 \times 20 \times 1 \times \log \frac{V_2}{1}$
$92.12 = 46.06 \log V_2$
$\log V_2 = \frac{92.12}{46.06} = 2$
$V_2 = 10^2 = 100 \ L$
Thus,$X = 100$.

(D) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ of a system is given by the equation: $\Delta U = Q - W$,where $Q$ is the heat exchanged and $W$ is the work done by the system.
If the process is adiabatic,there is no exchange of heat,so $Q = 0$.
Substituting this into the equation gives $\Delta U = -W$.
This implies that the change in internal energy is equal to the work done on the system (or the negative of the work done by the system) during an adiabatic process.
Therefore,the change in internal energy equals adiabatic work.

(D) For an isothermal reversible expansion,the work done by the gas is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$.
Since the process is isothermal,$p_1 V_1 = nRT$.
Given $p_1 = 20 \ atm$ and $V_1 = 1.5 \ L$,we have $nRT = 20 \ atm \times 1.5 \ L = 30 \ L \ atm$.
Substituting the values into the work formula:
$W = -30 \times \ln(\frac{15}{1.5})$
$W = -30 \times \ln(10)$
$W = -30 \times 2.303 \times \log_{10}(10)$
$W = -30 \times 2.303 \times 1 = -69.09 \ L \ atm$.
The negative sign indicates that work is done by the system on the surroundings.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Given: Heat absorbed $q = +100 \ J$.
Work done on the gas $w = -P_{ext} \Delta V$.
$P_{ext} = 1.5 \ atm$,$\Delta V = V_f - V_i = 2.0 \ L - 8.0 \ L = -6.0 \ L$.
$w = -(1.5 \ atm) \times (-6.0 \ L) = +9.0 \ L \cdot atm$.
Convert work to Joules: $w = 9.0 \times 101.32 \ J = 911.88 \ J$.
Therefore,$\Delta U = 100 \ J + 911.88 \ J = 1011.88 \ J \approx 1011.9 \ J$.

(C) According to the first law of thermodynamics,$\Delta E = q + W$.
For an adiabatic process,there is no heat exchange with the surroundings,so $q = 0$.
Substituting $q = 0$ into the equation,we get $\Delta E = W$.
Since $W$ is the work done on the system,for expansion,the work done by the system is $W_{sys} = -W$.
Thus,$\Delta E = -W_{sys}$,which implies $\Delta W = -\Delta E$.

(A) The relationship between enthalpy change and internal energy change is given by: $\Delta_{r} H = \Delta_{r} U + \Delta n_{g} RT$.
Given: $\Delta_{r} H = x \ kJ \ mol^{-1}$,$T = 1000 \ K$,$R = 8.3 \ J \ mol^{-1} \ K^{-1} = 8.3 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$.
For the reaction $ABO_{3(s)} \rightarrow AO_{(s)} + BO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = (n_{g})_{products} - (n_{g})_{reactants} = 1 - 0 = 1$.
Substituting the values: $x = \Delta_{r} U + (1 \times 8.3 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1} \times 1000 \ K)$.
$x = \Delta_{r} U + 8.3$.
Therefore,$\Delta_{r} U = x - 8.3 \ kJ \ mol^{-1}$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by: $\Delta H = \Delta U + \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species: $\Delta n_g = (n_{products, g}) - (n_{reactants, g})$.
For the reaction $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$,the gaseous moles are: $\Delta n_g = 3 - (1 + 5) = 3 - 6 = -3$.
Given: $\Delta H = -2800 \ kJ \ mol^{-1}$,$T = 300 \ K$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values: $\Delta U = \Delta H - \Delta n_g RT = -2800 - (-3 \times 8.314 \times 10^{-3} \times 300)$.
$\Delta U = -2800 + 7.48 = -2792.52 \ kJ \ mol^{-1}$.
Note: If the reaction was assumed to have $\Delta n_g = 0$,the answer would be $-2800 \ kJ \ mol^{-1}$ (Option $B$).

(B) $C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$
$\Delta n_{g} = \text{Number of gaseous products} - \text{number of gaseous reactants}$
$\Delta n_{g} = 1 - \frac{1}{2} = 0.5$
$R = 8.314 \ J \cdot K^{-1} \ mol^{-1}$
$T = 298 \ K$
$\Delta H = \Delta U + \Delta n_{g} RT$
$\Delta H - \Delta U = \Delta n_{g} RT$
$\Delta H - \Delta U = 0.5 \times 8.314 \times 298 = 1238.786 \ J \cdot mol^{-1}$
$\text{The approximate value is } 1238 \ J \cdot mol^{-1}$.

(C) $(C) \because$ According to the first law of thermodynamics,the total energy of an isolated system remains constant,although it may change from one form to another.

(A) From the first law of thermodynamics,$\Delta U = Q + W$.
Since heat is provided to the system,$Q = +50 \ J$.
Since work is done on the system,$W = +10 \ J$.
Therefore,the change in internal energy is $\Delta U = 50 \ J + 10 \ J = 60 \ J$.

(A) The reaction is $H_2O_{(l)} \rightarrow H_2O_{(s)}$ at $0^{\circ}C$ and $1 \ atm$.
Since the change in volume $(\Delta V)$ for the phase transition of water at $0^{\circ}C$ is negligible,the work done $(P\Delta V)$ is approximately zero.
According to the first law of thermodynamics,$\Delta H = \Delta U P\Delta V$.
Since $P\Delta V \approx 0$,we have $\Delta H \approx \Delta U$.
Given $\Delta U = -41 \ kJ / mol$,therefore $\Delta H = -41 \ kJ / mol$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by $\Delta H = \Delta E + \Delta n_g RT$.
For $\Delta H \neq \Delta E$,the change in the number of gaseous moles $(\Delta n_g)$ must not be equal to $0$.
$(A)$ $\Delta n_g = 1 - 1 = 0$.
$(B)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(C)$ $\Delta n_g = 2 - (1 + 1) = 0$.
$(D)$ $\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Since $\Delta n_g \neq 0$ for option $D$,$\Delta H \neq \Delta E$ for this reaction.

(C) For an adiabatic process,the heat exchange $dq = 0$.
Since the gas expands against a vacuum,the external pressure $P_{ext} = 0$,so the work done $dw = -P_{ext} \times dV = 0$.
According to the first law of thermodynamics,$dU = dq + dw$.
Substituting the values,$dU = 0 + 0 = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only $(U = f(T))$,so if $\Delta U = 0$,then $\Delta T = 0$.
Therefore,the correct statement is $\Delta U = 0$.

(B) The process involves heating water from $5^{\circ}C$ to $100^{\circ}C$ and then converting it into steam: $H_2O(\ell)_{5^{\circ}C} \longrightarrow H_2O(\ell)_{100^{\circ}C} \rightleftharpoons H_2O(g)_{100^{\circ}C}$.
Since heat is absorbed by the system from the microwave,$q = +ve$.
As the water expands during boiling against the external pressure,the system does work on the surroundings,so $w = -ve$.
The internal energy of the system increases because the temperature rises and the phase changes from liquid to gas,which has higher potential energy,thus $\Delta U = +ve$.

(D) For the process $X \to Y$: The change in internal energy is $\Delta U_1 = Q_1 - W_1 = 10 \text{ J} - 18 \text{ J} = -8 \text{ J}$.
Since internal energy is a state function,for the return process $Y \to X$,the change in internal energy is $\Delta U_2 = -\Delta U_1 = 8 \text{ J}$.
In the return process,$6 \text{ J}$ of heat is evolved,so $Q_2 = -6 \text{ J}$.
Using the first law of thermodynamics,$\Delta U_2 = Q_2 - W_2$,we have $8 \text{ J} = -6 \text{ J} - W_2$.
Solving for $W_2$,we get $W_2 = -6 \text{ J} - 8 \text{ J} = -14 \text{ J}$.
The negative sign indicates that work is done on the gas by the surroundings. Therefore,$14 \text{ J}$ of work is done on the gas '$A$' by the surroundings.

According to the first law of thermodynamics, $\Delta U = q + w$. Here, heat absorbed by the system $q = +500 \text{ J}$, and work done by the system $w = -200 \text{ J}$. Thus, $\Delta U = 500 - 200 = 300 \text{ J}$.