Basic concepts Questions in English

(B) The reaction is $4 HCl_{(g)} + O_{2_{(g)}} \rightarrow 2 Cl_{2_{(g)}} + 2 H_{2}O_{(g)}$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g = (n_{products} - n_{reactants}) = (2 + 2) - (4 + 1) = 4 - 5 = -1$.
This $\Delta n_g$ is for $4 \ moles$ of $HCl$.
For $1 \ mole$ of $HCl$,$\Delta n_g = -1 / 4 = -0.25$.
The work done is given by $W = -\Delta n_g RT$.
Given $T = 27^{\circ} C = 300 \ K$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
$W = -(-0.25) \times 8.314 \times 300$.
$W = 0.25 \times 2494.2 = 623.55 \ J \approx 623.6 \ J$.

(B) The work done by the system during expansion is given by the formula: $W = -P_{ext} \Delta V$.
Given:
External pressure $P_{ext} = 2 \times 10^5 \ Nm^{-2}$
Change in volume $\Delta V = 200 \ cm^3 = 200 \times 10^{-6} \ m^3 = 2 \times 10^{-4} \ m^3$.
Substituting the values:
$W = -(2 \times 10^5 \ Nm^{-2}) \times (2 \times 10^{-4} \ m^3)$
$W = -40 \ J$.
Therefore,the work done by the system is $-40 \ J$.

(D) According to the first law of thermodynamics,$\Delta U = q + w$.
For a process at constant volume (isochoric process),the work done $w = -P_{ext} \cdot \Delta V = 0$.
Therefore,$\Delta U = q_v$,where $q_v$ is the heat of reaction at constant volume.

(C) - $(1)$ Ratio of pressure to volume: This is incorrect. The ratio of pressure to volume is not related to enthalpy.
- $(2)$ Product of pressure and volume: This is also incorrect. The product of $P$ and $V$ is part of the enthalpy formula but does not define it by itself.
- $(3)$ Internal energy $(U)$ + $PV$: This is the correct definition of enthalpy. Enthalpy $(H)$ is defined as the sum of internal energy $(U)$ and the product of pressure $(P)$ and volume $(V)$,i.e.,$H = U + PV$.
- $(4)$ Internal energy $(U)$ - $PV$: This is incorrect. Enthalpy involves adding,not subtracting,$PV$ from internal energy.

(C) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 1 \ bar$,$V_1 = 3 \ L$,$V_2 = 15 \ L$.
Change in volume $\Delta V = V_2 - V_1 = 15 \ L - 3 \ L = 12 \ L$.
Since $1 \ L \ bar = 100 \ J$,we have $\Delta V = 12 \ dm^3$.
$W = -1 \ bar \times (15 \ L - 3 \ L) = -12 \ L \ bar$.
Converting to Joules: $W = -12 \ L \ bar \times 100 \ \frac{J}{L \ bar} = -1200 \ J = -1.200 \times 10^3 \ J$.

(A) The work done in an irreversible isothermal expansion against constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given:
$V_1 = 300 \ cm^3 = 0.3 \ dm^3$ (since $1 \ dm^3 = 1000 \ cm^3$).
$V_2 = 2.5 \ dm^3$.
$P_{ext} = 1.9 \ bar$.
$\Delta V = V_2 - V_1 = 2.5 \ dm^3 - 0.3 \ dm^3 = 2.2 \ dm^3$.
$W = -1.9 \ bar \times 2.2 \ dm^3 = -4.18 \ bar \cdot dm^3$.
Since $1 \ bar \cdot dm^3 = 100 \ J$,
$W = -4.18 \times 100 \ J = -418 \ J$.

(A) The internal energy of a substance increases with an increase in temperature due to the increase in rotational,translational,and vibrational energy of the molecules.

(B) Given: $n = 2$ moles,$V_1 = 5$ $dm^3$,$V_2 = 10$ $dm^3$,$P_{ext} = 1.5$ bar.
The formula for work done in an irreversible isothermal expansion against constant external pressure is $W = -P_{ext} \Delta V$.
$\Delta V = V_2 - V_1 = 10$ $dm^3 - 5$ $dm^3 = 5$ $dm^3$.
$W = -1.5 \text{ bar} \times 5$ $dm^3 = -7.5$ $dm^3$ bar.
Thus,the work done is $-7.5$ $dm^3$ bar.

(A) For an isothermal reversible compression,the work done is given by the formula: $W = -2.303 \ nRT \log_{10} \frac{V_2}{V_1}$
Given: $n = 2 \ mol$,$T = 300 \ K$,$V_1 = 40 \ L$,$V_2 = 20 \ L$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
Substituting the values: $W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} \frac{20}{40}$
$W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} (0.5)$
Since $\log_{10} (0.5) \approx -0.3010$:
$W = -2.303 \times 2 \times 8.314 \times 300 \times (-0.3010) \approx 3457.97 \ J$
Converting to $kJ$: $W \approx 3.46 \ kJ$
Since work is done on the system,the value is positive.

(B) Isochoric Process:
In an isochoric process,the change in volume of the thermodynamic system is zero.
Since the volume change is zero,the work done is also zero.
- Volume of the system = Constant
- Change in volume $\Delta V = 0$
- If $\Delta V = 0$,then work done $W = P \Delta V = 0$.
- According to the $1^{st}$ law of thermodynamics:
- $Q = \Delta U + W$
- Since $W = 0$,$Q = \Delta U$.

(D) Given: $V_1 = 2.5 \ L$,$V_2 = 4.5 \ L$,$W = -500 \ J$.
Since $100 \ J = 1 \ L \ bar$,then $W = -500 \ J = -5 \ L \ bar$.
The formula for work done in expansion is $W = -P_{ext} \Delta V$.
Substituting the values: $-5 \ L \ bar = -P_{ext} \times (4.5 \ L - 2.5 \ L)$.
$-5 \ L \ bar = -P_{ext} \times (2.0 \ L)$.
$P_{ext} = \frac{5 \ L \ bar}{2.0 \ L} = 2.5 \ bar$.

(A) For a reversible isothermal process,the work done $W$ is given by $W = -2.303 \ nRT \ \log_{10}\left(\frac{P_1}{P_2}\right)$.
Given: $n = 2 \ mol$,$T = 27 + 273 = 300 \ K$,$P_1 = 5.05 \times 10^6 \ Nm^{-2}$,$P_2 = 1.01 \times 10^5 \ Nm^{-2}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}\left(\frac{5.05 \times 10^6}{1.01 \times 10^5}\right)$.
$W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10}(50)$.
$W = -2.303 \times 2 \times 8.314 \times 300 \times 1.699 \approx -19520 \ J = -19.52 \ kJ$.
The negative sign indicates work is done by the system during expansion,but the question asks for work done $ON$ the system during compression. Note: The pressure values provided imply expansion $(P_1 > P_2)$. If compressed,$P_2 > P_1$. Assuming the magnitude is requested: $19.52 \ kJ$.

(B) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 2 \ atm$,$\Delta V = 1.5 \ L$.
Substituting the values: $W = -2 \ atm \times 1.5 \ L = -3 \ atm \cdot L$.
Since $1 \ atm \cdot L = 101.325 \ J$,we have: $W = -3 \times 101.325 \ J = -303.975 \ J$.
Rounding to one decimal place,we get $W \cong -303.9 \ J$.

(D) The relationship between $\Delta H$ and $\Delta U$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging this,we get $\Delta H - \Delta U = \Delta n_g RT$.
For the reaction $2 C_{(s)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (n_{g, \text{products}}) - (n_{g, \text{reactants}})$.
Here,$n_{g, \text{products}} = 1$ (for $C_2H_{6(g)}$) and $n_{g, \text{reactants}} = 3$ (for $H_{2(g)}$).
Thus,$\Delta n_g = 1 - 3 = -2$.
Substituting this value into the equation,we get $\Delta H - \Delta U = -2 RT$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta (PV)$.
For ideal gases,this can be written as $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species.
In solids and liquids,the change in volume $(\Delta V)$ is very small,making the term $P\Delta V$ negligible.
However,for systems consisting of gases,the change in volume is significant,making the difference between $\Delta H$ and $\Delta U$ substantial.
Therefore,the difference is significant for systems consisting of only gases.

(A) reversible process is one that proceeds infinitely slowly such that the system remains in equilibrium with its surroundings at every stage.
In a reversible process,the driving force and the opposing force differ by an infinitesimal amount,not a large amount.
Therefore,the statement that 'The driving and opposing forces differ by a large amount' is $NOT$ a feature of a reversible process.

(C) The work done in a chemical reaction at constant pressure is given by the formula: $W = -\Delta n_g RT$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_g$:
$\Delta n_g = (n_{products}) - (n_{reactants}) = (2 + 2) - (4 + 1) = 4 - 5 = -1$.
Now,substitute the values into the work formula:
$W = -(-1) \times 8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K$.
$W = 1 \times 8.314 \times 300 \ J = 2494.2 \ J$.
Rounding to the nearest integer,the work done is $2494 \ J$.

(A) free expansion means expansion against zero opposing force,so $P_{ext} = 0$,which implies $W = 0$.
For an isothermal process of an ideal gas,the internal energy change $\Delta U = 0$.
The enthalpy change is given by the formula $\Delta H = \Delta U + \Delta(PV)$.
Since $\Delta U = 0$ and for an ideal gas at constant temperature $\Delta(PV) = 0$,it follows that $\Delta H = 0 \ kJ$.

(B) An isothermal process is defined as a process in which the temperature of the system remains constant $(dT = 0)$.
For an ideal gas,the internal energy $(U)$ is a function of temperature only $(U = f(T))$. Therefore,if the temperature is constant,the internal energy remains constant $(dU = 0)$.
However,for real gases or condensed phases,enthalpy $(H = U + PV)$ depends on pressure as well. Even if temperature is constant,enthalpy may change if the pressure changes.
Thus,the statement that the enthalpy of the system remains constant is not universally true for all isothermal processes,making it the false statement.

(C) The work done in a $PV$ system is given by the formula $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Initial volume $V_1 = 150 \ mL + 150 \ mL = 300 \ mL = 0.3 \ dm^3$.
Final volume $V_2 = 150 \ mL = 0.15 \ dm^3$.
External pressure $P_{ext} = 1 \ bar$.
Substituting the values: $W = -1 \ bar \times (0.15 \ dm^3 - 0.3 \ dm^3) = -1 \times (-0.15) \ dm^3 \ bar = 0.15 \ dm^3 \ bar$.
Since $1 \ dm^3 \ bar = 100 \ J$,the work done is $0.15 \times 100 \ J = 15.00 \ J$.

(B) Thermodynamic properties are classified as either state functions or path functions.
State functions depend only on the initial and final states of the system,such as $Internal \ energy$ $(U)$,$Enthalpy$ $(H)$,and $Entropy$ $(S)$.
Path functions depend on the path taken to reach the final state,such as $Work$ $(w)$ and $Heat$ $(q)$.
Therefore,$Work$ is a path function.

(D) The work done in a chemical reaction is given by the formula $W = -\Delta n_{g} RT$.
First,calculate the change in the number of moles of gaseous species,$\Delta n_{g} = n_{p(g)} - n_{r(g)}$.
For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)}$,$\Delta n_{g} = 1 - (1 + 0.5) = -0.5 \ mol$.
Now,substitute the values into the work formula: $W = -(-0.5) \times 8.314 \times 300$.
$W = 0.5 \times 8.314 \times 300 = 1247.1 \ J$.
Note: If the question asks for work done $BY$ the system,it is $1247.1 \ J$. If it asks for work done $ON$ the system,it is $-1247.1 \ J$. Given the options,the magnitude is $1247.1 \ J$.

(D) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$
Given:
$P_{ext} = 2.5 \ bar$
$V_1 = 2.5 \ L$
$V_2 = 4.5 \ L$
Calculation:
$\Delta V = V_2 - V_1 = 4.5 \ L - 2.5 \ L = 2.0 \ L$
$W = -2.5 \ bar \times 2.0 \ L = -5.0 \ bar \cdot L$
Since $1 \ bar \cdot L = 100 \ J$:
$W = -5.0 \times 100 \ J = -500 \ J$

(A) For an ideal gas,the internal energy $(U)$ is a function of temperature only,i.e.,$U = f(T)$.
Since the process is isothermal,the temperature remains constant,so $\Delta T = 0$.
Therefore,the change in internal energy $\Delta U = nC_v\Delta T = 0$.

(A) state function is a property whose value depends only on the current state of the system and not on the path taken to reach that state. $P$,$V$,and $T$ are state functions. Work $(w)$ and heat $(q)$ are path functions,meaning their values depend on the process or path taken. Therefore,the statement that work is a state function is false.

(C) The work done during the compression of a gas at constant pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here, $P_{ext} = 2 \ bar$, $V_1 = 10 \ dm^3$, and $V_2 = 1 \ dm^3$.
$\Delta V = V_2 - V_1 = 1 \ dm^3 - 10 \ dm^3 = -9 \ dm^3$.
$W = -2 \ bar \times (-9 \ dm^3) = +18 \ bar \ dm^3$.
Since $1 \ bar \ dm^3 = 100 \ J$, then $18 \ bar \ dm^3 = 1800 \ J = 1.8 \ kJ$.
Since the work is done on the gas, the value is positive.

(C) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
First,convert the volumes to the same unit $(m^3)$:
$V_1 = 5 \ dm^3 = 5 \times 10^{-3} \ m^3$.
$V_2 = 7 \times 10^{-3} \ m^3$.
Change in volume: $\Delta V = V_2 - V_1 = (7 \times 10^{-3} - 5 \times 10^{-3}) \ m^3 = 2 \times 10^{-3} \ m^3$.
Now,calculate the work done:
$W = -(2.02 \times 10^5 \ Nm^{-2}) \times (2 \times 10^{-3} \ m^3)$.
$W = -4.04 \times 10^2 \ J = -404 \ J$.
Converting to $kJ$: $W = -0.404 \ kJ$.

(B) The formula for work done against constant external pressure is $W = -P_{ext} \Delta V$.
First,convert the volumes to the same unit $(m^3)$:
$V_1 = 2 \ dm^3 = 2 \times 10^{-3} \ m^3$.
$V_2 = 6 \times 10^{-3} \ m^3$.
Change in volume $\Delta V = V_2 - V_1 = (6 - 2) \times 10^{-3} \ m^3 = 4 \times 10^{-3} \ m^3$.
Given external pressure $P_{ext} = 1 \ bar = 10^5 \ Pa = 10^5 \ N/m^2$.
Substituting the values:
$W = -(10^5 \ N/m^2) \times (4 \times 10^{-3} \ m^3) = -400 \ J$.
Converting to $kJ$:
$W = -400 / 1000 \ kJ = -0.4 \ kJ$.

(A) process in which the volume of the system remains constant throughout the change is known as an $Isochoric$ process.
Therefore,for an $Isochoric$ process,the change in volume $\Delta V = 0$.

(A) Internal energy $(U)$ is a state function.
Therefore,the change in internal energy $(\Delta U)$ depends only on the initial and final states of the system,not on the path taken.

(A) Work of compression occurs when the volume of the system decreases during a reaction,which corresponds to a negative change in the number of moles of gas $(\Delta n_g < 0)$.
In option $A$: $NH_{3(g)} + HCl_{(g)} \longrightarrow NH_4Cl_{(s)}$,the change in moles of gas is $\Delta n_g = 0 - (1 + 1) = -2$. Since the volume decreases significantly as gaseous reactants form a solid,this reaction shows work of compression.
In option $B$: $\Delta n_g = 3 - 1 = +2$ (Expansion).
In option $C$: $\Delta n_g = (2 + 1) - 2 = +1$ (Expansion).
In option $D$: $\Delta n_g = 1 - 0 = +1$ (Expansion).
Therefore,the correct reaction is $NH_{3(g)} + HCl_{(g)} \longrightarrow NH_4Cl_{(s)}$.

(D) The formula for work done during compression is $W = -P_{ext} \Delta V = -P_{ext}(V_2 - V_1)$.
Given $P_{ext} = 4.05 \ bar$,$V_1 = 2.5 \times 10^{-2} \ m^3$,and $V_2 = 1.3 \times 10^{-2} \ m^3$.
Substituting the values: $W = -4.05 \ bar \times (1.3 \times 10^{-2} - 2.5 \times 10^{-2}) \ m^3$.
$W = -4.05 \times (-1.2 \times 10^{-2}) \ bar \cdot m^3 = 4.86 \times 10^{-2} \ bar \cdot m^3$.
Since $1 \ bar = 10^5 \ Pa$ and $1 \ Pa \cdot m^3 = 1 \ J$,we convert the units:
$W = 4.86 \times 10^{-2} \times 10^5 \ J = 4860 \ J$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{ext} \cdot \Delta V$.
Given $W = -6 \ dm^3 \ bar$ and $\Delta V = V_2 - V_1 = (20 - 15) \ dm^3 = 5 \ dm^3$.
Substituting the values: $-6 \ dm^3 \ bar = -P_{ext} \cdot (5 \ dm^3)$.
$P_{ext} = \frac{6}{5} \ bar = 1.2 \ bar$.
Since $1 \ bar = 10^5 \ Pa$,the external pressure is $1.2 \times 10^5 \ Pa$.

(D) For an isothermal reversible expansion process,the work done is given by the formula:
$W = -2.303 \ nRT \log \frac{P_1}{P_2}$
Given:
$n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 210 \ kPa$,$P_2 = 105 \ kPa$
Substituting the values:
$W = -2.303 \times 1 \times 8.314 \times 300 \times \log \frac{210}{105}$
$W = -2.303 \times 8.314 \times 300 \times \log 2$
Using $\log 2 \approx 0.3010$:
$W = -2.303 \times 8.314 \times 300 \times 0.3010$
$W \approx -1729 \ J$
The magnitude of work done is $|W| = 1729 \ J$.

(A) The expansion of a gas against no external pressure (opposing force),such as expansion into a vacuum,is defined as $free \ expansion$.

(A) The work done during expansion is given by the formula $W = -P_{ext} \Delta V$.
Since the gas expands,work is done by the system,so $W = -5 \ bar \ dm^3$.
Given $P_{ext} = 2.5 \ bar$,$V_2 = 4.5 \ L$,and $V_1 = x \ L$.
Since $1 \ L = 1 \ dm^3$,we have:
$-5 \ bar \ dm^3 = -2.5 \ bar \times (4.5 \ L - x \ L)$.
Dividing both sides by $-2.5 \ bar$:
$2 = 4.5 - x$.
Rearranging for $x$:
$x = 4.5 - 2 = 2.5 \ L$.

(A) The formula for work done during expansion against constant external pressure is $W = -P_{\text{ext}} \Delta V$.
Given: $W = -5.09 \ kJ = -5090 \ J$,$V_1 = 2 \times 10^{-2} \ m^3$,$V_2 = 3 \times 10^{-2} \ m^3$.
Change in volume $\Delta V = V_2 - V_1 = (3 - 2) \times 10^{-2} \ m^3 = 1 \times 10^{-2} \ m^3$.
Substituting the values: $-5090 = -P_{\text{ext}} \times (1 \times 10^{-2})$.
$P_{\text{ext}} = \frac{5090}{10^{-2}} = 509000 \ Nm^{-2} = 5.09 \times 10^5 \ Nm^{-2}$.

(A) Given: $n = 2 \ mol$,$V_{1} = 12.5 \ L = 12.5 \times 10^{-3} \ m^{3}$,$V_{2} = 15.0 \ L = 15.0 \times 10^{-3} \ m^{3}$,$P_{ext} = 760 \ mm \ Hg = 1.013 \times 10^{5} \ N \ m^{-2}$.
Work done in an irreversible expansion is given by the formula: $W = -P_{ext} \Delta V$.
$W = -P_{ext}(V_{2} - V_{1}) = -1.013 \times 10^{5} \times (15.0 - 12.5) \times 10^{-3} \ J$.
$W = -1.013 \times 10^{5} \times 2.5 \times 10^{-3} \ J$.
$W = -1.013 \times 250 \ J$.
$W = -253.25 \ J$.

(D) Given: $V_1 = 10 \ dm^3, V_2 = 100 \ dm^3, T = 300 \ K, R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Mass of $O_2$ $(m)$ = $1.6 \times 10^{-2} \ kg = 16 \ g$.
Molar mass of $O_2$ $(M)$ = $32 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{16}{32} = 0.5 \ mol$.
For an isothermal and reversible expansion process,the work done is given by:
$W = -2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$.
Substituting the values:
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log_{10} \left( \frac{100}{10} \right)$.
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log_{10} (10)$.
Since $\log_{10} (10) = 1$,
$W = -2.303 \times 0.5 \times 8.314 \times 300 = -2872 \ J$.

(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ex} \Delta V$.
Given:
$V_{1} = 1 \times 10^{-3} \ m^{3} = 0.001 \ m^{3}$
$V_{2} = 1 \times 10^{-2} \ m^{3} = 0.01 \ m^{3}$
$P_{ex} = 1 \times 10^{5} \ Nm^{-2}$
Change in volume $\Delta V = V_{2} - V_{1} = 0.01 - 0.001 = 0.009 \ m^{3}$.
Substituting the values:
$W = -1 \times 10^{5} \times 0.009 = -900 \ J = -9 \times 10^{2} \ J$.

(A) The work done during the compression of a gas against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given:
$P_{ext} = 100 \ kPa = 10^{5} \ Pa$
$V_{initial} = 5 \ m^{3}$
$V_{final} = 2.5 \ m^{3}$
$\Delta V = V_{final} - V_{initial} = 2.5 \ m^{3} - 5 \ m^{3} = -2.5 \ m^{3}$
Substituting the values:
$W = -(10^{5} \ Pa) \times (-2.5 \ m^{3})$
$W = 2.5 \times 10^{5} \ J$
$W = 250,000 \ J = 250 \ kJ$.
Since the gas is compressed, work is done on the system, resulting in a positive value.

(D) Given: $n = 2 \text{ mol}$,$V_{1} = 5 \text{ m}^{3}$,$V_{2} = 10 \text{ m}^{3}$,$T = 300 \text{ K}$,$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log_{10} \frac{V_{2}}{V_{1}}$.
Substituting the values: $W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} \frac{10}{5}$.
$W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} 2$.
Since $\log_{10} 2 \approx 0.3010$,we get: $W = -2.303 \times 2 \times 8.314 \times 300 \times 0.3010$.
$W \approx -3457.97 \text{ J} = -3.458 \text{ kJ}$.

(A) The formula for work done against constant external pressure is $W = -P_{\text{ext}} \Delta V$.
Given: $P_{\text{ext}} = 100 \ kPa = 100 \times 10^3 \ Pa$,$V_i = 1 \ m^3$,$V_f = 10 \ dm^3 = 0.01 \ m^3$.
Change in volume $\Delta V = V_f - V_i = 0.01 \ m^3 - 1 \ m^3 = -0.99 \ m^3$.
$W = -(100 \times 10^3 \ Pa) \times (-0.99 \ m^3) = 99,000 \ J = +99 \ kJ$.
Since the gas is being compressed,work is done on the system,hence the value is positive.

(D) Given,pressure $p = 100 \ kPa = 10^{5} \ Pa$.
Initial volume $V_{1} = 1 \ m^{3}$.
Final volume $V_{2} = 10 \ dm^{3} = 10 \times 10^{-3} \ m^{3} = 0.01 \ m^{3}$.
Work done on the system is given by $W = -p_{ext} \Delta V$.
Since the gas is compressed,work is done on the system,so $W = p_{ext}(V_{1} - V_{2})$.
$W = 10^{5} \ Pa \times (1 \ m^{3} - 0.01 \ m^{3}) = 10^{5} \times 0.99 \ J = 99,000 \ J$.

(B) An endothermic process is one that absorbs heat from the surroundings.
Phase transitions that involve moving from a more ordered state to a less ordered state (e.g.,solid to liquid or liquid to gas) require energy input to overcome intermolecular forces.
In the transformation $H_2O_{(s)} \rightarrow H_2O_{(\ell)}$,ice melts into liquid water,which is an endothermic process because heat is absorbed to break the crystal lattice of ice.
The other options ($H_2O_{(\ell)} \rightarrow H_2O_{(s)}$,$H_2O_{(g)} \rightarrow H_2O_{(\ell)}$,and $H_2O_{(g)} \rightarrow H_2O_{(s)}$) represent freezing or condensation,which are exothermic processes as they release heat.

(B) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta(PV)$.
At constant volume,$\Delta V = 0$.
Since $\Delta V = 0$,the work done $P \Delta V = 0$.
Therefore,$\Delta H = \Delta U$,which implies $\Delta H - \Delta U = 0$.

(B) By definition,the standard enthalpy of formation of any element in its most stable state at standard conditions ($298 \ K$ and $1 \ bar$) is defined as zero.
Since dihydrogen $(H_2)$ is an element in its standard state,its standard enthalpy of formation is $0 \ kJ/mol$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging this equation,we get: $\Delta H - \Delta U = \Delta n_g RT$.
For the given reaction: $C_3H_{8(g)} + 5O_{2(g)} \longrightarrow 3CO_{2(g)} + 4H_2O_{(l)}$.
$\Delta n_g$ is the difference between the number of moles of gaseous products and gaseous reactants.
$\Delta n_g = (n_{products, g}) - (n_{reactants, g}) = 3 - (1 + 5) = 3 - 6 = -3$.
Substituting the value of $\Delta n_g$ into the equation: $\Delta H - \Delta U = -3RT$.

(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 reactions involving only solids or liquids,there is no change in the number of moles of gas,so $\Delta n_g = 0$.
Therefore,$\Delta H = \Delta U + (0)RT$,which simplifies to $\Delta H = \Delta U$.

(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: $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$.
The change in the number of moles of gaseous species is calculated as: $\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$.
$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$.
Substituting this value into the equation,we get: $\Delta H = \Delta U + (-1) RT$.
Therefore,$\Delta H = \Delta U - RT$.