Basic concepts Questions in English

(D) For an ideal gas,the enthalpy $H$ is a function of temperature only,i.e.,$H = f(T)$.
Since the process is isothermal,the temperature remains constant,so $\Delta T = 0$.
The change in enthalpy is given by $\Delta H = n C_p \Delta T$.
Since $\Delta T = 0$,the enthalpy change $\Delta H$ is $0 \ kJ$.

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

(C) We know that $H = E + PV$.
For an ideal gas,$PV = nRT$.
Therefore,$H = E + nRT$.
For two different states at constant temperature $T$: $H_1 = E_1 + n_1RT$ and $H_2 = E_2 + n_2RT$.
Subtracting the first equation from the second: $(H_2 - H_1) = (E_2 - E_1) + (n_2RT - n_1RT)$.
Rearranging this gives: $H_2 - H_1 - E_2 + E_1 = n_2RT - n_1RT$.

(C) For an ideal gas,the heat of reaction at constant pressure $(q_{p})$ and constant volume $(q_{v})$ are related by the enthalpy change equation.
Since $\Delta H = \Delta E + \Delta n RT$ and $\Delta H = q_{p}$ (at constant pressure) and $\Delta E = q_{v}$ (at constant volume),
Therefore,$q_{p} = q_{v} + \Delta n RT$.

(D) The properties of a system whose values are independent of the amount of substance present in the system are called intensive properties.
Examples include $temperature$,$viscosity$,$surface \ tension$,$pressure$,$density$,$molar \ heat \ capacity$,etc.
Since all the given options are independent of the amount of matter,they are all intensive properties.

(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 $\Delta H = \Delta U$ to hold true,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 the reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$:
$\Delta n_g = (1 + 1) - 2 = 0$.
Therefore,for this reaction,$\Delta H = \Delta U$.

(D) Given: $n = 2 \ mol$,$V_1 = 1 \ L$,$V_2 = 10 \ L$,$T = 300 \ K$,$R = 0.0083 \ kJ \ K^{-1} \ mol^{-1}$.
For a reversible isothermal expansion,the work done is given by the formula:
$W = -2.303 \ nRT \log \frac{V_2}{V_1}$.
Substituting the values:
$W = -2.303 \times 2 \times 0.0083 \times 300 \times \log \frac{10}{1}$.
$W = -2.303 \times 2 \times 0.0083 \times 300 \times 1$.
$W = -11.47 \ kJ \approx -11.5 \ kJ$.
The magnitude of work done by the system is $11.5 \ kJ$.

(A) The work done during expansion is given by the formula $W = -P_{ext} \Delta V$.
Since the gas is expanding against an external pressure,the work done by the gas is $W = P_{ext} \Delta V$.
Given $P_{ext} = 10^{5} \ Nm^{-2}$,$V_{1} = 1 \ m^{3}$,and $V_{2} = 2 \ m^{3}$.
$\Delta V = V_{2} - V_{1} = 2 \ m^{3} - 1 \ m^{3} = 1 \ m^{3}$.
$W = 10^{5} \ Nm^{-2} \times 1 \ m^{3} = 10^{5} \ J$.
Converting to kilojoules,$10^{5} \ J = 100 \ kJ = 10^{2} \ kJ$.

(C) The enthalpy change $(\Delta H)$ is related to the internal energy change $(\Delta U)$ by the equation:
$\Delta H = \Delta U + \Delta(PV)$
For gaseous reactions,this is expressed as:
$\Delta H = \Delta U + \Delta n_{g}RT$
where $\Delta n_{g}$ is the change in the number of moles of gaseous products and reactants.

(D) $I$) Work is a path function because it depends on the path taken to reach the final state from the initial state. This is a correct statement.
$II$) Enthalpy $(H)$ is an extensive property because it depends on the amount of matter present in the system. This is a correct statement.
$III$) Lattice enthalpy of ionic compounds can be calculated using the Born-Haber cycle,which relates lattice enthalpy to other thermodynamic data like ionization energy,electron gain enthalpy,and sublimation energy. This is a correct statement.
Therefore,all statements $I, II,$ and $III$ are correct.

(D) Extensive properties depend on the amount of matter present in the system. These include: enthalpy,mass,and volume. (Total = $3$)
Intensive properties are independent of the amount of matter present in the system. These include: density,temperature,and pressure. (Total = $3$)
Therefore,the number of extensive and intensive properties is $3$ and $3$ respectively.

(C) Extensive properties are those that depend on the amount of matter present in the system.
From the given list:
$1$. Enthalpy $(H)$: Extensive property.
$2$. Density $(d)$: Intensive property (ratio of mass to volume).
$3$. Volume $(V)$: Extensive property.
$4$. Internal energy $(U)$: Extensive property.
$5$. Temperature $(T)$: Intensive property.
Therefore,the extensive properties are enthalpy,volume,and internal energy.
The total number of extensive properties is $3$.

(C) Extensive properties depend on the amount of matter present in the system,such as $Internal \ energy$,$enthalpy$,$Mass$,$volume$,$Heat \ capacity$,and $Gibbs \ energy$.
Intensive properties are independent of the amount of matter present,such as $Density$ and $pressure$.
Therefore,the pair in which both are not extensive properties (i.e.,both are intensive) is $Density$ and $pressure$.

(C) Extensive properties depend on the amount of matter present in the system. These are: $\text{Mass}$,$\text{Enthalpy}$,$\text{Heat capacity}$,and $\text{Internal energy}$. (Total = $4$)
Intensive properties are independent of the amount of matter present in the system. These are: $\text{Temperature}$,$\text{Pressure}$,and $\text{Density}$. (Total = $3$)
Therefore,the number of extensive and intensive properties is $4$ and $3$ respectively.

(A) Extensive properties are those that depend on the quantity or size of the matter present in the system. Their values change if the amount of matter changes.
Among the given properties:
$1$. Volume: Extensive
$2$. Enthalpy: Extensive
$3$. Density: Intensive (ratio of mass to volume)
$4$. Temperature: Intensive
$5$. Heat capacity: Extensive
$6$. Pressure: Intensive
$7$. Internal energy: Extensive
The extensive properties are Volume,Enthalpy,Heat capacity,and Internal energy.
Therefore,the total number of extensive properties is $4$.

(A) Extensive properties are those properties whose value depends on the quantity or size of matter present in the system.
$(A)$ Heat capacity: Extensive property.
$(B)$ Entropy: Extensive property.
$(C)$ Gibbs' energy: Extensive property.
$(D)$ Concentration: Intensive property (independent of the amount of matter).
$(E)$ Vapour pressure: Intensive property (independent of the amount of matter).
Therefore,$A, B$ and $C$ are extensive properties.

(B) State functions are thermodynamic properties that depend only on the initial and final states of the system,not on the path taken to reach that state.
$Internal \ Energy$,$Enthalpy$,and $Entropy$ are state functions.
$Work$ and $Heat$ are path functions,meaning their values depend on the process path taken.

(D) Extensive properties are those that depend on the amount of matter or mass present in the system.
$I$. Molar entropy is an intensive property because it is defined per mole.
$II$. Specific volume is an intensive property because it is defined per unit mass.
$III$. Heat capacity is an extensive property as it depends on the total mass of the substance.
$IV$. Volume is an extensive property as it depends on the total amount of matter.
Therefore,$III$ and $IV$ are extensive properties.

(B) Extensive properties are properties that depend on the amount of matter present in the system,such as volume,mass,and enthalpy.
Intensive properties are independent of the amount of matter present,such as temperature,pressure,and density.
$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{extensive property}}{\text{extensive property}} = \text{Intensive property}$.
Therefore,volume is an extensive property.

(D) State functions are path independent,such as enthalpy $(H)$,internal energy $(U)$,and free energy $(G)$.
Work $(W)$ is a path-dependent function and is not a state function.

(C) For an ideal gas,the enthalpy $H$ is a function of temperature only,i.e.,$H = f(T)$.
Since the process is isothermal,the change in temperature $\Delta T = 0$.
Therefore,the change in enthalpy $\Delta H = nC_p\Delta T = 0$.

(C) In a $P-V$ diagram,the work done during a thermodynamic process is given by the area under the curve projected onto the volume axis ($V$-axis).
For the process $B \rightarrow E$,the work done is the area enclosed by the curve $BE$,the vertical lines $BC$ and $ED$,and the volume axis $CD$.
This corresponds to the area of the region $BCDE$ under the curve $BE$.

(C) The general relation between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by:
$\Delta H = \Delta U + \Delta n_g RT$
For $1 \text{ mole}$ of an ideal gas undergoing a process where $\Delta n_g = 1$,the equation becomes:
$\Delta H = \Delta U + RT$
Rearranging this to solve for $\Delta U$:
$\Delta U = \Delta H - RT$
Since $\Delta T$ is the change in temperature,for a process where the change in moles is $1$,the term is $R \Delta T$. Thus,$\Delta U = \Delta H - R \Delta T$.

(B) The formula for work done in isothermal reversible expansion is $W_{rev} = -2.303 nRT \log \left(\frac{V_2}{V_1}\right)$.
Given: $n = 2 \ mol$,$V_1 = 5 \ L$,$V_2 = 50 \ L$,$W = -189.1 \ L \ atm$,and $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
Substituting the values: $-189.1 = -2.303 \times 2 \times 0.082 \times T \times \log \left(\frac{50}{5}\right)$.
$-189.1 = -2.303 \times 2 \times 0.082 \times T \times 1$.
$T = \frac{-189.1}{-2.303 \times 2 \times 0.082} \approx 500 \ K$.
Converting to Celsius: $T(^{\circ}C) = 500 - 273 = 227^{\circ}C$.

(C) The work done by the system during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \times (V_2 - V_1)$.
Given: $P_{ext} = 2 \ atm$,$V_1 = 5 \ L$,$V_2 = X \ L$,and $W = -2,026.4 \ J$.
First,convert the work from Joules to $L \cdot atm$ using the conversion factor $1 \ L \cdot atm = 101.32 \ J$:
$W = \frac{-2,026.4 \ J}{101.32 \ J \cdot L^{-1} \cdot atm^{-1}} = -20 \ L \cdot atm$.
Now,substitute the values into the work formula:
$-20 \ L \cdot atm = -2 \ atm \times (X - 5) \ L$.
Dividing both sides by $-2 \ atm$:
$10 = X - 5$.
Therefore,$X = 15 \ L$.

(A) According to the first law of thermodynamics,$\Delta U = q + W$.
For an adiabatic process,there is no heat exchange,so $q = 0$. Thus,$\Delta U = W_{\text{adiabatic}}$. Statement $I$ is correct.
Work is a path function because the amount of work done depends on the specific path taken between two states,not just the initial and final states. Statement $II$ is correct.
An extensive property is a property whose value depends on the quantity or size of matter present in the system. Volume is directly proportional to the amount of substance,making it an extensive property. Statement $III$ is correct.
Therefore,all statements $I, II,$ and $III$ are correct.

(B) For an ideal gas,the equation of state is $PV = nRT$.
Since the pressure $P$ is constant,the change in volume $\Delta V$ due to a change in temperature $\Delta T$ is given by $P \Delta V = nR \Delta T$.
The work done $W$ in a process at constant pressure is defined as $W = P \Delta V$.
Substituting the expression for $P \Delta V$,we get $W = nR \Delta T$.
Given $n = 2$ moles and $\Delta T = 20 \ K$ (or $20^{\circ} C$),the work done is $W = 2 \times R \times 20 = 40R$.

(B) The pressure $P = 10^7 \ N \cdot m^{-2}$.
Initial volume $V_i = 5 \ m^3$.
Final volume $V_f = 10 \ m^3$.
Change in volume $\Delta V = V_f - V_i = 10 \ m^3 - 5 \ m^3 = 5 \ m^3$.
The work done on the gas during expansion against constant external pressure is given by $W = -P_{ext} \Delta V$.
Substituting the values: $W = -(10^7 \ N \cdot m^{-2}) \times (5 \ m^3) = -50 \times 10^6 \ J$.
Since $1 \ MJ = 10^6 \ J$,we get $W = -50 \ MJ$.

(B) An adiabatic process is defined as a thermodynamic process in which there is no exchange of heat between the system and its surroundings.
Therefore,for an adiabatic change,the heat exchange $(q)$ is equal to zero $(q = 0)$.

(C) The work done by a gas is given by the formula $W = p \Delta V$.
An isochoric process is defined as a process in which the volume remains constant,which implies $\Delta V = 0$.
Substituting this into the work formula,we get $W = p \times 0 = 0$.
Therefore,the work done by an ideal gas at a constant volume is $0$.
Hence,the correct option is $(C)$.

(B) For an isothermal reversible expansion of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log(\frac{V_f}{V_i})$.
Since $PV = nRT$,we can substitute $nRT$ with $P_i V_i$ (initial state):
$w = -2.303 \ P_i V_i \ \log(\frac{V_f}{V_i})$.
Given: $w = -184.24 \ L \ atm$,$P_i = 10 \ atm$,$V_i = 4 \ L$.
Substituting the values:
$-184.24 = -2.303 \times (10 \times 4) \times \log(\frac{V_f}{4})$.
$-184.24 = -2.303 \times 40 \times \log(\frac{V_f}{4})$.
$-184.24 = -92.12 \times \log(\frac{V_f}{4})$.
$\log(\frac{V_f}{4}) = \frac{-184.24}{-92.12} = 2$.
$\frac{V_f}{4} = 10^2 = 100$.
$V_f = 100 \times 4 = 400 \ L$.

(D) The formula for enthalpy change during heating is $\Delta H = n C_p \Delta T$.
Given:
$C_p = 75 \ J \ mol^{-1} \ K^{-1}$
$T_1 = 10^{\circ}C = 283 \ K$
$T_2 = 20^{\circ}C = 293 \ K$
$\Delta T = 293 - 283 = 10 \ K$
Number of moles $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{9 \ g}{18 \ g \ mol^{-1}} = 0.5 \ mol$.
Substituting the values:
$\Delta H = 0.5 \ mol \times 75 \ J \ mol^{-1} \ K^{-1} \times 10 \ K$
$\Delta H = 375 \ J$.

(A) Internal energy $(U)$ is a state function.
State functions depend only on the initial and final states of the system and are independent of the path taken to reach the final state.
In both processes $I$ and $II$,the system starts at state $A$ and ends at state $B$.
Therefore,the change in internal energy for both processes is the same: $\Delta U_1 = U_B - U_A$ and $\Delta U_2 = U_B - U_A$.
Hence,$\Delta U_1 = \Delta U_2$.

(A) Intensive properties are those properties of a system that do not depend on the quantity or size of matter present in the system.
Analyzing the given list:
$I$. Molar volume: Intensive (volume per mole).
$II$. Mass: Extensive.
$III$. Internal energy: Extensive.
$IV$. Volume: Extensive.
$V$. Enthalpy: Extensive.
$VI$. Temperature: Intensive.
$VII$. Density: Intensive (mass per unit volume).
Therefore,the intensive properties are $I, VI,$ and $VII$.

(A) An intensive property is a physical quantity whose value does not depend on the amount of the substance for which it is measured.
For example,the melting point of water is $0^{\circ} C$,whether you take $100 \ mL$ of water or $1 \ kg$ of water.
Other examples of intensive properties are specific gravity and refractive index.
Entropy is an extensive property,not an intensive property,because its value depends on the amount of substance present.
As the amount of substance increases,the entropy also increases.

(D) An extensive property is a property of matter that depends on the amount of matter present in the system. Examples include volume,entropy,and heat capacity at constant volume.
Conversely,an intensive property is independent of the amount of matter present. Molar heat capacity at constant pressure is defined for $1 \ mol$ of a substance,making it an intensive property.
Therefore,the correct answer is $(D)$.

(B) state function is a property of a system that depends only on its initial and final states,not on the path taken to reach that state.
Internal energy $(U)$,entropy $(S)$,and free energy $(G)$ are examples of state functions.
Work $(w)$ depends on the path followed by the system during a process,therefore it is a path function.

(A) For an isothermal reversible expansion of an ideal gas,the work done $(w)$ is given by the formula: $w = -nRT \ln(\frac{P_1}{P_2})$.
Given: $n = 1 \ mol$,$T = 300 \ K$,$P_1 = 20 \ atm$,$P_2 = 2 \ atm$,and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $w = -1 \times 8.3 \times 300 \times \ln(\frac{20}{2})$.
$w = -2490 \times \ln(10)$.
Using $\ln(10) \approx 2.303$: $w = -2490 \times 2.303 = -5734.47 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $w = -5.734 \ kJ \ mol^{-1}$.
Since the work done is $-x \ kJ \ mol^{-1}$,we have $-x = -5.734$,so $x = 5.73$.

(A) Given: $n = 1 \ mol$,$T = 61 \ K$,$V_1 = 1.0 \ L$,$V_2 = 10.0 \ L$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$.
For an isothermal reversible expansion,the work done is given by the formula:
$W = -nRT \ln(V_2 / V_1)$
Substituting the values into the equation:
$W = -(1) \times (0.0821) \times (61) \times \ln(10 / 1)$
$W = -5.0081 \times 2.303$
$W \approx -11.53 \ L \ atm$.
Rounding to the nearest provided option,the work done is $-11.52 \ L \ atm$.

(D) For a reversible isothermal expansion,the work done is given by the formula: $W = -nRT \ln \frac{P_1}{P_2} = -2.303 nRT \log \frac{P_1}{P_2}$.
Given: $n = 2 \ mol$,$P_1 = 10 \ atm$,$P_2 = 1 \ atm$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values: $W = -2.303 \times 2 \times 8.3 \times T \times \log \frac{10}{1}$.
Since $\log 10 = 1$,we get $W = -2.303 \times 2 \times 8.3 \times T \times 1 = -38.2298 \times T \ J$.
To convert to $kJ$,divide by $1000$: $W = -38.2298 \times 10^{-3} \times T \ kJ \approx -3.82 \times 10^{-2} \times T \ kJ$.

(B) Work done is equal to the area under the curve in a $P-V$ diagram.
In the given graph,the area under the path represents the work done by the gas.
Comparing the areas under the three paths,the area under path $3$ is the largest,followed by path $2$,and the area under path $1$ is the smallest.
Therefore,the work done follows the relation $W_1 < W_2 < W_3$.

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + P\Delta V$.
For solids and liquids,the change in volume $(\Delta V)$ is extremely small,making the term $P\Delta V$ negligible.
Therefore,for solids and liquids,$\Delta H \approx \Delta U$.
In contrast,for gases,$\Delta V$ is significant,and $P\Delta V = \Delta n_g RT$,which is not negligible.
Thus,the difference is not significant for $(i)$ Solids and $(iv)$ Liquids.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + P\Delta V$.
At $0^{\circ} C$ and $1 \ bar$,the change in volume $(\Delta V)$ for the freezing of $1 \ g$ of water is extremely small and negligible.
Therefore,$\Delta H \approx \Delta U$.
Given that the enthalpy change for freezing $1 \ g$ of water is $334 \ J$,the internal energy change is also approximately $334 \ J$.

(C) The absolute value of enthalpy $(H)$ cannot be determined experimentally. Only the change in enthalpy $(\Delta H)$ during a process can be measured. Therefore,statement $C$ is incorrect.

(A, B, C) An extensive property depends on the amount of matter,while an intensive property is independent of the amount of matter.
$1$. The product of an intensive property $(x)$ and an extensive property $(X)$ is extensive ($xX$ is extensive).
$2$. The ratio of an intensive property $(x)$ to an extensive property $(X)$ is intensive ($\frac{x}{X}$ is intensive).
$3$. The ratio of an extensive property $(X)$ to an intensive property $(x)$ is extensive ($\frac{X}{x}$ is extensive).
$4$. The ratio of the change in an extensive property $(dX)$ to the change in an intensive property $(dx)$ is extensive ($\frac{dX}{dx}$ is extensive).
Therefore,statements $A$,$B$,and $C$ are correct.

(A, C, D) Extensive properties are those whose values depend on the quantity or size of matter present in the system.
$H$ (Enthalpy),$E$ (Internal energy),and $V$ (Volume) are all extensive properties because they scale with the amount of substance.
$p$ (Pressure) is an intensive property as it is independent of the amount of matter.
Therefore,$H$,$E$,and $V$ are extensive variables.

(C) According to the first law of thermodynamics,$Q = \Delta U + W$.
In an adiabatic process,$Q = 0$.
In free expansion,the external pressure $P_{ext} = 0$,so the work done $W = P_{ext} \Delta V = 0$.
Substituting these values into the first law: $0 = \Delta U + 0$,which implies $\Delta U = 0$.
For an ideal gas,internal energy $U$ is a function of temperature only $(U = f(T))$.
Since $\Delta U = 0$,the change in temperature $\Delta T = 0$,meaning the temperature remains constant.
Therefore,the process is isothermal.

(A) The Gibbs free energy is defined as $G = H - TS$.
Taking the differential,we get $dG = dH - TdS - SdT$.
Since $H = U + PV$,the differential is $dH = dU + PdV + VdP$.
From the first law of thermodynamics,$dU = TdS - PdV$.
Substituting $dU$ into the expression for $dH$,we get $dH = (TdS - PdV) + PdV + VdP = TdS + VdP$.
Now,substituting $dH$ back into the expression for $dG$:
$dG = (TdS + VdP) - TdS - SdT$.
Simplifying this,we obtain $dG = VdP - SdT$.