First law of thermodynamics Questions in English

(C) Given: Heat absorbed $(Q)$ = $+200 \ J$ (since heat is absorbed by the system).
Change in volume $(\Delta V)$ = $500 \ cm^3 = 500 \times 10^{-6} \ m^3 = 5 \times 10^{-4} \ m^3$.
External pressure $(P_{ext})$ = $2 \times 10^5 \ N \ m^{-2}$.
Work done $(W)$ = $-P_{ext} \times \Delta V$.
$W = -(2 \times 10^5 \ N \ m^{-2}) \times (5 \times 10^{-4} \ m^3) = -100 \ J$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
$\Delta U = 200 \ J + (-100 \ J) = +100 \ J$.

(D) For an isobaric process,the pressure remains constant $( P = \text{constant} )$.
According to the first law of thermodynamics,$ \Delta U = Q + W $.
The work done in an expansion or compression process is given by $ W = -P_{ext} \Delta V $.
Substituting this into the first law equation: $ \Delta U = Q_P - P_{ext} \Delta V $.
Rearranging the terms gives: $ Q_P = \Delta U + P_{ext} \Delta V $.

(B) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that heat is supplied to the system,$Q = +40 \ J$.
Since the system does work on the surroundings,$W = -8 \ J$.
Substituting these values into the equation:
$\Delta U = 40 \ J + (-8 \ J) = 32 \ J$.
Therefore,the increase in internal energy is $32 \ J$.

(NONE) The chemical equation for the oxidation of $4 \ mol$ of $SO_{2(g)}$ is:
$4 SO_{2(g)} + 2 O_{2(g)} \longrightarrow 4 SO_{3(g)}$
Calculate the change in the number of moles of gaseous species,$\Delta n_g$:
$\Delta n_g = (n_{product, gas}) - (n_{reactant, gas}) = 4 - (4 + 2) = -2 \ mol$
The work done $(W)$ is given by the formula:
$W = -\Delta n_g RT$
Given $T = 27^{\circ} C = 300 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$:
$W = -(-2 \ mol) \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K$
$W = +4988.4 \ J = +4.988 \ kJ$
Since the calculated value is $+4.988 \ kJ$ and it does not match any of the given options,the provided options are incorrect.

(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{g} RT$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$,the change in the number of gaseous moles is $\Delta n_{g} = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = -0.5$.
Substituting this into the equation,we get $\Delta H = \Delta U - 0.5 RT$.
Since $R$ and $T$ are positive,$-0.5 RT$ is negative,therefore $\Delta H < \Delta U$.

(B) Given: Heat absorbed $(q)$ = $+150 \ J$ (since heat is absorbed by the system).
External pressure $(P_{ext})$ = $2 \times 10^5 \ N \ m^{-2}$.
Change in volume $(\Delta V)$ = $300 \ cm^3 = 300 \times 10^{-6} \ m^3 = 3 \times 10^{-4} \ m^3$.
Work done $(W)$ = $-P_{ext} \times \Delta V$.
$W = -(2 \times 10^5 \ N \ m^{-2}) \times (3 \times 10^{-4} \ m^3) = -60 \ J$.
According to the first law of thermodynamics:
$\Delta U = q + W$.
$\Delta U = 150 \ J + (-60 \ J) = 90 \ J$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Since the gas performs work on the surroundings,$w = -900 \ J$.
The internal energy increases,so $\Delta U = +625 \ J$.
Substituting these values: $625 = q - 900$,which gives $q = 1525 \ J$.
For an ideal gas,the enthalpy change is defined as $\Delta H = \Delta U + P\Delta V$.
Since $w = -P\Delta V$,we have $P\Delta V = -w = 900 \ J$.
Therefore,$\Delta H = 625 \ J + 900 \ J = 1525 \ J$.

(A) In an adiabatic process,there is no exchange of heat between the system and its surroundings,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + W$.
For adiabatic compression,work is done on the system,so $W > 0$.
Therefore,$\Delta U = W > 0$,which means the internal energy increases.
In isothermal processes,$\Delta T = 0$,so $\Delta U = 0$ for an ideal gas.

(B) Given: Heat absorbed $(Q)$ = $+210 \ J$.
External pressure $(P_{ext})$ = $10^5 \ Pa = 1 \ bar$.
Change in volume $(\Delta V)$ = $6 \ L - 3 \ L = 3 \ L = 3 \ dm^3$.
Work done $(W)$ = $-P_{ext} \times \Delta V = -1 \ bar \times 3 \ dm^3 = -3 \ L \cdot bar$.
Since $1 \ L \cdot bar = 100 \ J$,then $W = -3 \times 100 \ J = -300 \ J$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
$\Delta U = 210 \ J + (-300 \ J) = -90 \ J$.

(C) The change in the number of moles of gaseous species is given by $\Delta n_g = (n_{products, g} - n_{reactants, g})$.
For the reaction: $NH_2CN_{(g)} + \frac{3}{2} O_{2_{(g)}} \longrightarrow N_{2_{(g)}} + CO_{2_{(g)}} + H_2O_{(l)}$
$\Delta n_g = (1 + 1) - (1 + 1.5) = 2 - 2.5 = -0.5 \ mol$.
Given $\Delta U = -740.5 \ kJ$,$T = 25 + 273 = 298 \ K$,and $R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Using the relation $\Delta H = \Delta U + \Delta n_g RT$:
$\Delta H = -740.5 \ kJ + (-0.5 \ mol) \times (8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) \times 298 \ K$.
$\Delta H = -740.5 \ kJ - 1.2388 \ kJ = -741.7388 \ kJ \ mol^{-1}$.
Rounding to one decimal place,we get $\Delta H \approx -741.7 \ kJ \ mol^{-1}$.

(B) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done on the system,so $w = +65 \ kJ$.
The system releases heat,so $q = -25 \ kJ$.
Substituting these values: $\Delta U = -25 \ kJ + 65 \ kJ = +40 \ kJ$.

(B) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 2 \ bar$,$V_1 = 0.2 \ dm^3$,$V_2 = 0.8 \ dm^3$.
Change in volume: $\Delta V = V_2 - V_1 = 0.8 \ dm^3 - 0.2 \ dm^3 = 0.6 \ dm^3$.
Substituting the values: $W = -2 \ bar \times 0.6 \ dm^3 = -1.2 \ bar \cdot dm^3$.
Since $1 \ bar \cdot dm^3 = 100 \ J$,we convert the units: $W = -1.2 \times 100 \ J = -120 \ J$.
Thus,the work done by the gas is $-120 \ J$.

(C) According to the $I^{st}$ law of thermodynamics:
$\Delta U = q + w$
Here,heat is released,so $q = -2 \ kJ$.
Work is done on the system,so $w = +10 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 10 \ kJ = 8 \ kJ$.

(D) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q + W$.
Since the system is exothermic,the heat released is $Q = -260 \ kJ$.
Since the system does work,the work done by the system is $W = -10 \ kJ$.
Therefore,$\Delta U = -260 \ kJ + (-10 \ kJ) = -270 \ kJ$.

(A) Since the container is insulated,the process is adiabatic,so $\delta Q = 0$.
According to the first law of thermodynamics,$\Delta U = \delta Q + \delta W$.
Here,$\delta W = -P_{ext} \times \Delta V$.
Given: $P_{ext} = 2.5 \ bar = 2.5 \times 10^5 \ Pa$,$V_1 = 4.5 \ dm^3 = 4.5 \times 10^{-3} \ m^3$,$V_2 = 7 \times 10^{-3} \ m^3$.
$\Delta V = (7 - 4.5) \times 10^{-3} \ m^3 = 2.5 \times 10^{-3} \ m^3$.
$\Delta U = 0 + [-(2.5 \times 10^5 \ Pa) \times (2.5 \times 10^{-3} \ m^3)]$.
$\Delta U = -6.25 \times 10^2 \ J = -625 \ J$.

(A) According to the $I^{st}$ law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the equation: $\Delta U = q + w$.
Here,work is done by the system,so $w = -37.6 \ J$.
Heat is lost by the system,so $q = -14.6 \ J$.
Substituting these values into the equation:
$\Delta U = -14.6 \ J + (-37.6 \ J) = -52.2 \ J$.

(C) According to the first law of thermodynamics,the change in internal energy ($\Delta U$ or $\Delta E$) is given by the equation: $\Delta E = q + w$.
Here,heat is added to the system,so $q = +40 \ kJ$.
The system does work on the surroundings,so $w = -140 \ kJ$.
Substituting these values into the equation: $\Delta E = 40 \ kJ + (-140 \ kJ) = -100 \ kJ$.

(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = q + w$.
When work is done on the system,$w = +X \ J$.
When heat is transferred to the surrounding,$q = -Y \ J$.
Therefore,$\Delta U = (-Y) + (+X) = X - Y \ J$.

(D) According to the first law of thermodynamics,the change in internal energy $(\Delta U)$ is given by the sum of heat $(q)$ added to the system and the work $(w)$ done on the system: $\Delta U = q + w$.
In this question,$Q$ is defined as the heat liberated from the system,so $q = -Q$.
$W$ is defined as the work done on the system,so $w = W$.
Substituting these into the equation: $\Delta U = -Q + W$.
Rearranging for $Q$: $Q = W - \Delta U$.

(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 $q = +30 \ kJ$.
The system performs work on the surroundings,so $w = -12 \ kJ$.
Substituting these values: $\Delta U = 30 \ kJ + (-12 \ kJ) = 18 \ kJ$.
Therefore,the increase in internal energy is $18 \ kJ$.

(D) For an insulated container,the process is adiabatic,so $q = 0$.
According to the first law of thermodynamics,$\Delta U = q + w$.
The work done during expansion against a constant external pressure is given by $w = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 2.5 \ atm$,$V_1 = 2.5 \ L$,and $V_2 = 4.5 \ L$.
$\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$.
$w = -2.5 \ atm \times 2.0 \ L = -5.0 \ atm \cdot L$.
Since $1 \ atm \cdot L = 101.3 \ J$,we have $w = -5.0 \times 101.3 \ J = -506.5 \ J$.
Therefore,$\Delta U = 0 + (-506.5 \ J) = -506.5 \ J$.

(A) The work done by a gas is given by $W = -P_{ext} \cdot \Delta V$.
Since the process occurs at constant volume,$\Delta V = 0$,which implies $W = 0$.
According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that the heat supplied $Q = x \ J$ and $W = 0$,we get $\Delta U = x \ J$.
Therefore,$Q = \Delta U = x \ J$ and $W = 0$.

(A) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that heat $Q$ is provided to the system,$Q = +x \ kJ = 1000x \ J$.
Since work $W$ is done on the system,$W = +y \ J$.
Therefore,the change in internal energy is $\Delta U = 1000x + y \ J$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
Heat absorbed by the system is positive,so $q = +54 \ J$.
Work done by the system on the surroundings is negative,so $w = -238 \ J$.
Substituting these values into the equation:
$\Delta U = 54 \ J + (-238 \ J) = -184 \ J$.
Thus,the change in internal energy of the system is $-184 \ J$.

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

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Since the system gives out heat,$q = -x \ J$.
Since the system does work on its surroundings,$w = -y \ J$.
Therefore,$\Delta U = (-x) + (-y) = -x - y \ J$.

(A) According to the $I^{st}$ law of thermodynamics,$\Delta U = q + w$.
Given: Heat absorbed by the system,$q = +710 \ J$.
Change in internal energy,$\Delta U = +460 \ J$.
Substituting these values into the equation: $460 \ J = 710 \ J + w$.
$w = 460 \ J - 710 \ J$.
$w = -250 \ J$.
Since the work done $w$ is negative,it indicates that work is done by the system.

(A) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
Since the work done is given by $W = -P \Delta V$,for an isochoric process,$W = 0$.
The first law of thermodynamics is expressed as $\Delta U = Q + W$.
Substituting $W = 0$ and $Q = Q_v$ (heat at constant volume),we get $\Delta U = Q_v$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Given: $P_{ext} = 2.40 \times 10^5 \ Pa$,$V_2 = 2.2 \times 10^{-3} \ m^3$,$W = -0.048 \ kJ = -48 \ J$.
Substituting the values: $-48 \ J = -2.4 \times 10^5 \ Pa \times (2.2 \times 10^{-3} \ m^3 - V_1)$.
Dividing both sides by $-2.4 \times 10^5 \ Pa$: $\frac{-48}{-2.4 \times 10^5} = 2.2 \times 10^{-3} - V_1$.
$20 \times 10^{-5} = 2.2 \times 10^{-3} - V_1$.
$0.2 \times 10^{-3} = 2.2 \times 10^{-3} - V_1$.
$V_1 = 2.2 \times 10^{-3} - 0.2 \times 10^{-3} = 2.0 \times 10^{-3} \ m^3$.

(D) According to the first law of thermodynamics,$\Delta U = q + W$.
Work done on the system is positive,so $W = +62 \ J$.
Heat transferred to the surrounding is negative,so $q = -128 \ J$.
Substituting these values into the equation:
$\Delta U = -128 \ J + 62 \ J = -66 \ J$.
Therefore,the internal energy change is $-66 \ J$.

(D) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that the system absorbs heat,$Q = +8 \ kJ$.
Since the system does work on the surroundings,$W = -2.2 \ kJ$.
Substituting these values into the equation: $\Delta U = 8 \ kJ + (-2.2 \ kJ) = 5.8 \ kJ$.
Therefore,the internal energy change is $5.8 \ kJ$.

(B) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given: Heat absorbed by the system,$Q = +4000 \ kJ$.
Work done on the system by the surroundings,$W = +2000 \ J = +2 \ kJ$.
Therefore,$\Delta U = 4000 \ kJ + 2 \ kJ = 4002 \ kJ$.

(C) According to the first law of thermodynamics,$\Delta U = q + w$.
Since the process occurs at constant volume,the work done $w = -P \Delta V = 0$.
Given that heat is supplied to the system,$q = +500 \ J$.
Therefore,$q = 500 \ J$ and $w = 0$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,work is done on the system,so $w = +2.5 \ kJ = +2500 \ J$.
The system releases heat,so $q = -1500 \ J$.
Therefore,$\Delta U = -1500 \ J + 2500 \ J = 1000 \ J$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
For an isobaric process (constant pressure),the heat exchanged is $q_p = \Delta H = \Delta U + P\Delta V$.
Since work done $w = -P\Delta V$,we have $P\Delta V = -w$.
Substituting this,$q_p = \Delta U - w$,which means $\Delta U = q_p + w$.
Thus,option $A$ is correct.
For an adiabatic process,$q = 0$,so $\Delta U = w$.
For an isochoric process,$\Delta V = 0$,so $w = 0$ and $\Delta U = q_v$.
For an isothermal process of an ideal gas,$\Delta U = 0$,so $q = -w$.

(C) According to the first law of thermodynamics,$\Delta U = q + W$.
Since the gas performs work on the surrounding,the work done is negative: $W = -0.320 \ kJ = -320 \ J$.
The heat absorbed by the system is positive: $q = 120 \ J$.
Substituting these values into the equation: $\Delta U = 120 \ J + (-320 \ J) = -200 \ J$.

(C) The statement “The mass and energy both are conserved in an isolated system” refers to the modified first law of thermodynamics.
According to the original first law of thermodynamics,energy can neither be created nor destroyed,only transformed.
However,considering the mass-energy equivalence principle $(E = mc^2)$,the law is modified to state that the total mass and energy of an isolated system remain constant.

(A) Given: $n = 2 \ mol$,$V_1 = 10 \ dm^3 = 10 \times 10^{-3} \ m^3 = 0.01 \ m^3$,$V_2 = 2 \ m^3$,$P_{ext} = 101.325 \ kPa = 101.325 \times 10^3 \ Pa$.
The formula for work done in an irreversible expansion is $W = -P_{ext} \cdot \Delta V$.
$\Delta V = V_2 - V_1 = 2 \ m^3 - 0.01 \ m^3 = 1.99 \ m^3$.
$W = -101.325 \times 10^3 \ Pa \times 1.99 \ m^3$.
$W = -201636.75 \ J = -201.63675 \ kJ$.
Rounding to one decimal place,$W = -201.6 \ kJ$.

(A) The first law of thermodynamics is given by the equation: $\Delta U = q + w$.
Under isobaric conditions,the work done is given by $w = -p_{ex} \cdot \Delta V$.
Substituting this into the first law equation,we get: $\Delta U = q - p_{ex} \cdot \Delta V$.

(A) The first law of thermodynamics is given by the equation: $\Delta U = q + W$.
For an isothermal process,the temperature remains constant,which implies that the internal energy change is zero,i.e.,$\Delta U = 0$.
Substituting this into the first law equation,we get $0 = q + W$,which simplifies to $q = -W$.

(D) The combustion reaction is: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$.
Work done is given by the formula: $w = -\Delta n_g RT$.
For $1 \ mol$ of $CH_4$,the change in gaseous moles is $\Delta n_g = n_{g(products)} - n_{g(reactants)} = 1 - (1 + 2) = -2$.
For $0.5 \ mol$ of $CH_4$,$\Delta n_g = 0.5 \times (-2) = -1 \ mol$.
Substituting the values: $w = -(-1 \ mol) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \times (300 \ K)$.
$w = +2494 \ J$.

(A) The first law of thermodynamics is given by the equation $\Delta U = q + W$.
For an isochoric process,the volume remains constant,which means the change in volume $\Delta V = 0$.
Since work done $W = -P_{ext} \Delta V$,it follows that $W = 0$.
Substituting $W = 0$ into the first law equation,we get $\Delta U = q_v$,where $q_v$ is the heat exchanged at constant volume.

(B) The formula for work done in an isothermal reversible expansion is $W = -2.303 \ nRT \log \frac{V_2}{V_1}$.
Given:
Mass of $O_2 = 16 \ g$,Molar mass of $O_2 = 32 \ g/mol$,so $n = \frac{16}{32} = 0.5 \ mol$.
$T = 300 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$V_1 = 10 \ dm^3$,$V_2 = 100 \ dm^3$.
Substituting the values:
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log \frac{100}{10}$
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log(10)$
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times 1$
$W \approx -2872.9 \ J \approx -2875 \ J$ (rounding to the nearest option).

(C) According to the first law of thermodynamics:
$\Delta U = q + w$
For an adiabatic process,there is no exchange of heat with the surroundings,so $q = 0$.
Substituting this into the equation,we get $\Delta U = w$.
This implies that the work done $(w)$ is equal to the change in internal energy $(\Delta U)$.
If work is done by the system $(w < 0)$,the internal energy decreases $(\Delta U < 0)$,meaning work is done at the expense of internal energy.

(D) For a chemical reaction,the heat of reaction at constant pressure is equal to the change in enthalpy,$\Delta H = q_p$.
The heat of reaction at constant volume is equal to the change in internal energy,$\Delta U = q_v$.
From the first law of thermodynamics,$\Delta H = \Delta U + \Delta(PV)$.
For an ideal gas,$PV = nRT$,so $\Delta(PV) = \Delta nRT$.
Therefore,the relationship is $\Delta H = \Delta U + \Delta nRT$.

(D) The relationship between enthalpy change $(\Delta H)$,internal energy change $(\Delta U)$,and work done $(W)$ is given by: $\Delta H = \Delta U + P \Delta V$.
Since work done by the system during expansion is $W = -P \Delta V$,we have $P \Delta V = -W$.
Therefore,$\Delta H = \Delta U - W$.
Given:
Internal energy change $\Delta U = +432 \ J$.
Work done by the gas $W = 200 \ J$.
Since the gas performs work,the work done on the system is $-200 \ J$.
Substituting the values: $\Delta H = 432 \ J - (-200 \ J) = 432 \ J + 200 \ J = 632 \ J$.

(D) According to the first law of thermodynamics,the change in internal energy ($\Delta U$ or $\Delta E$) is given by the equation: $\Delta U = q + w$.
Here,the system gains heat,so $q = +x \ J$.
Work is done on the system,so $w = +y \ J$.
Therefore,$\Delta U = (+x) + (+y) = x + y \ J$.

(C) The first law of thermodynamics is given by $\Delta U = q + w$.
Since $\Delta H = q_p$ (heat at constant pressure),we have $\Delta H = \Delta U - w$.
Given: $\Delta U = 600 \ J$ and $w = -800 \ J$.
Substituting these values: $\Delta H = 600 - (-800) = 1400 \ J$.
Wait,if the work done is on the system,$w$ is positive. If work is done by the system,$w$ is negative.
Using $\Delta H = \Delta U + P \Delta V$,where $P \Delta V = -w$.
$\Delta H = 600 + (-800) = -200 \ J$.

(C) The balanced chemical equation for the formation of $2$ moles of ammonia is:
$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$
We know the relation $\Delta H - \Delta U = \Delta n_g RT$.
Here,$\Delta n_g$ is the change in the number of moles of gaseous species:
$\Delta n_g = (\text{Sum of stoichiometric coefficients of gaseous products}) - (\text{Sum of stoichiometric coefficients of gaseous reactants})$
$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting the value of $\Delta n_g$ in the relation:
$\Delta H - \Delta U = -2 RT$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H - \Delta U = \Delta n_g RT$
Here,$\Delta n_g$ is the change in the number of moles of gaseous species.
For the reaction: $2 \ C_6H_{6(\ell)} + 15 \ O_{2(g)} \rightarrow 12 \ CO_{2(g)} + 6 \ H_2O_{(\ell)}$
$\Delta n_g = (n_{g, \text{products}}) - (n_{g, \text{reactants}}) = 12 - 15 = -3$
Now,substituting the values: $\Delta H - \Delta U = -3 \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 298 \ K$
$= -7432.7 \ J = -7.432 \ kJ$