Basic concepts Questions in English

(D) Statement $(1)$ is correct: The Joule-Thomson process occurs under adiabatic conditions $(q = 0)$ and is isoenthalpic $(dH = 0)$.
Statement $(2)$ is correct: The Joule-Thomson coefficient is defined as $\mu_{JT} = (\partial T / \partial P)_H$. If $\mu_{JT} < 0$,then for expansion $(dP < 0)$,$dT$ must be positive,meaning the gas warms up.
Statement $(3)$ is correct: The inversion temperature $(T_i)$ is the temperature at which the Joule-Thomson coefficient $\mu_{JT}$ is zero,meaning no temperature change occurs upon expansion.
Therefore,all three statements are correct.

(D) In a $Cyclic \ process$,a system undergoes a series of changes through different states,but ultimately returns to its initial state.
Since the initial and final states are identical,the change in any state function (like internal energy,enthalpy,etc.) for a complete cycle is zero.

(C) Internal energy $(U)$ is a state function.
For any cyclic process,the change in a state function is zero.
Since the internal energy change $(\Delta U)$ depends only on the initial and final states,and for a cycle the initial and final states are the same,$\Delta U = 0$.

(C) In an isolated system,neither the exchange of matter nor the exchange of energy is possible with the surroundings. Since a well-stoppered thermos flask prevents both heat transfer and matter exchange,it acts as an isolated system.

(D) . Intensive properties are those that do not depend on the quantity or size of matter present in the system.
$Temperature$ and $refractive \ index$ are independent of the amount of substance,hence they are intensive properties.
$Enthalpy$ and $volume$ are extensive properties because they depend on the amount of matter present.

(C) In thermochemistry,the standard state of a substance is defined as its pure form at a pressure of $1 \,atm$ and a specified temperature,which is conventionally taken as $298.15 \,K$ or $25^{\circ}C$.

(A) For an ideal gas,internal energy $(E)$ and enthalpy $(H)$ are functions of temperature only.
Since the process is isothermal,the temperature remains constant $(\Delta T = 0)$.
Therefore,$\Delta E = 0$ and $\Delta H = 0$,meaning $E$ and $H$ remain unchanged.

(B) Enthalpy $(H)$ is defined as $H = U + PV$.
Since internal energy $(U)$,pressure $(P)$,and volume $(V)$ are state functions,enthalpy is also a state function.
State functions are independent of the path followed.
Enthalpy is an extensive property because its value depends on the amount of substance present in the system.
Therefore,the statement that it is an intensive property is incorrect.

(B) The stability of a system is inversely proportional to its energy content.
Therefore,a system with lower energy is more stable.

(B) The functions whose value depends only on the state of a system and not on the path taken to reach that state are known as state functions.
Internal energy $(U)$ is a state function because it depends only on the initial and final states of the system.

(D) In an adiabatic process,the system is thermally insulated from its surroundings,meaning no heat exchange occurs between the system and the surroundings. Therefore,$q = 0$.

(B) An intensive property is a physical property of a system that does not depend on the amount of matter present or the size of the system.
$Enthalpy$,$Mass$,and $Volume$ are extensive properties because they depend on the quantity of matter.
$Density$ is defined as $Mass / Volume$,which is an intensive property because the ratio of two extensive properties is intensive.

(C) An intensive property is a property that does not depend on the amount of matter present in the system.
Pressure,density,and temperature are intensive properties because they remain constant regardless of the size of the sample.
Volume is an extensive property because it depends on the amount of matter present in the system.
Therefore,the correct answer is $(C)$.

(C) An isolated system is defined as a system that does not allow the exchange of either energy (heat) or matter with its surroundings.

(D) $ \Delta Q $ (heat) and $ \Delta W $ (work) are path functions,not state functions.
State functions depend only on the initial and final states of the system,such as $ \Delta S $ (entropy change),$ \Delta G $ (Gibbs free energy change),and $ \Delta H $ (enthalpy change).

(C) In an adiabatic process,there is no exchange of heat between the system and the surroundings.
Therefore,the change in heat,$\Delta Q$,is equal to $0$.

(C) An intensive property is a physical property of a system that does not depend on the amount of matter or the size of the system.
$Mass$,$Volume$,and $Enthalpy$ are extensive properties because they depend on the quantity of matter.
$Surface \ tension$ is an intensive property as it remains constant regardless of the amount of substance present.
Therefore,the correct option is $(C)$.

(A) In an exothermic reaction,heat is released into the surroundings.
This implies that the enthalpy of the products $(H_P)$ is less than the enthalpy of the reactants $(H_R)$.
Mathematically,the change in enthalpy is given by $\Delta H = H_P - H_R$.
Since $H_P < H_R$,the value of $\Delta H$ is always negative for exothermic reactions.

(B) The enthalpy change $(\Delta H)$ is defined as the change in internal energy $(\Delta E)$ plus the pressure-volume work $(P \Delta V)$.
Therefore,the correct relation is $\Delta H = \Delta E + P \Delta V$.

(C) The fundamental relation between enthalpy $(\Delta H)$ and internal energy $(\Delta E)$ for a process at constant pressure is given by $\Delta H = \Delta E + P \Delta V$.
Rearranging this equation gives $\Delta E = \Delta H - P \Delta V$.
For gaseous reactions,the work done $P \Delta V$ is equivalent to $\Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous species. Thus,$\Delta H = \Delta E + \Delta n_g RT$ (often written as $\Delta H = \Delta E + n RT$).
Therefore,the relation $\Delta H = \Delta E - P \Delta V$ is incorrect.

(A) The law of $Lavoisier$ and $Laplace$ states that the amount of heat required to decompose a compound into its elements is equal to the heat evolved during the formation of that compound from its elements. This is a direct consequence of the principle of conservation of energy.

(C) For an ideal gas,the internal energy $(U)$ and enthalpy $(H)$ are functions of temperature only.
During an isothermal process,the temperature remains constant,so $\Delta T = 0$.
Since $\Delta H = nC_p\Delta T$,if $\Delta T = 0$,then $\Delta H = 0$.
Therefore,the enthalpy remains unaffected.

(A) By convention,the standard enthalpy of formation of an element in its most stable state at $298 \ K$ $(25 \ ^oC)$ and $1 \ \text{bar}$ pressure is defined as $Zero$.

(C) According to the first law of thermodynamics,at constant volume,the heat exchanged $(Q_v)$ is equal to the change in internal energy ($\Delta E$ or $\Delta U$).
Therefore,$Q_v = \Delta E = E_P - E_R$,where $E_P$ is the internal energy of the products and $E_R$ is the internal energy of the reactants.

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

(A) The internal energy $(U)$ of a substance is a state function that depends on the temperature of the system.
For most substances,as the temperature increases,the kinetic energy of the particles increases,which leads to an increase in the total internal energy of the substance.
Therefore,the internal energy increases with an increase in temperature.

(A) The first law of thermodynamics states that $\Delta E = q + w$.
For a process at constant pressure,the heat exchanged is equal to the change in enthalpy,i.e.,$q = \Delta H$.
Substituting this into the equation,we get $\Delta E = \Delta H + w$.
Rearranging this gives $\Delta H = \Delta E - w$.
However,in many conventions,work done by the system is represented as $W = -P\Delta V$,leading to $\Delta H = \Delta E + P\Delta V$.
If $W$ is defined as the work done on the system $(W = P\Delta V)$,then the relation is $\Delta H = \Delta E + W$.

(C) For an isothermal reversible expansion of an ideal gas,the work done $W$ is given by the formula:
$W = 2.303 \, nRT \, log_{10} \left( \frac{P_1}{P_2} \right)$
Given:
$n = 1 \, \text{mole}$,$R = 2 \, \text{cal} \, \text{K}^{-1} \, \text{mol}^{-1}$,$T = 300 \, \text{K}$,$P_1 = 10 \, \text{atm}$,$P_2 = 1 \, \text{atm}$.
Substituting the values:
$W = 2.303 \times 1 \times 2 \times 300 \times log_{10} \left( \frac{10}{1} \right)$
$W = 2.303 \times 600 \times 1 = 1381.8 \, \text{cal}$.

(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 $\Delta H = \Delta E$,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 $C$: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$.
Therefore,for this reaction,$\Delta H = \Delta E$.

(D) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$,which can be rearranged as $\Delta H - \Delta E = \Delta n_g RT$.
For the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the change in the number of gaseous moles is $\Delta n_g = (1 + 3) - 2 = 2$.
The temperature $T = 27 \ ^\circ C = 27 + 273 = 300 \ K$.
Substituting these values,we get $\Delta H - \Delta E = 2 \times R \times 300 = 8.314 \times 300 \times 2$.
Thus,the correct option is $D$.

(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 $\Delta H = \Delta E$,the term $\Delta n_g$ must be equal to $0$.
$\Delta n_g$ is the difference between the number of moles of gaseous products and gaseous reactants.
For option $C$: $H_{2(g)} + Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$.
Therefore,for this reaction,$\Delta H = \Delta E$.

(C) The work done during the isothermal reversible expansion of an ideal gas is given by the formula:
$W = 2.303 \, nRT \log_{10} \frac{P_1}{P_2}$
Given values:
$n = 1 \, mol$,$T = 300 \, K$,$P_1 = 10 \, atm$,$P_2 = 1 \, atm$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$
Substituting these values into the equation:
$W = 2.303 \times 1 \times 8.314 \times 300 \times \log_{10} \left( \frac{10}{1} \right)$
$W = 2.303 \times 8.314 \times 300 \times 1$
$W = 5744.1 \, J$
Thus,the correct option is $C$.

(C) For an ideal gas,the internal energy $(U)$ is a function of temperature only,i.e.,$U = f(T)$.
Since the process occurs at constant temperature $(T)$,the change in internal energy $(\Delta U)$ is zero.
Therefore,the internal energy remains the same.

(C) Enthalpy $(H)$ is a thermodynamic property defined as the sum of the internal energy $(E)$ and the product of pressure $(P)$ and volume $(V)$ of the system.
Mathematically,it is expressed as: $H = E + PV$.

(A) The internal energy $(U)$ of a system is a state function,which means its value depends only on the current state of the system and not on the path taken to reach that state.
Therefore,option $(A)$ is correct.

(C) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Here,$P_{ext} = 1 \times 10^5 \ N \ m^{-2}$,$V_1 = 1 \times 10^{-3} \ m^3$,and $V_2 = 1 \times 10^{-2} \ m^3$.
$\Delta V = V_2 - V_1 = (1 \times 10^{-2} - 1 \times 10^{-3}) \ m^3 = (10 \times 10^{-3} - 1 \times 10^{-3}) \ m^3 = 9 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(1 \times 10^5 \ N \ m^{-2}) \times (9 \times 10^{-3} \ m^3) = -900 \ J$.
Thus,the work done is $-900 \ J$.

(A) The exact value of internal energy is not known as it includes all types of energies of molecules constituting the given mass of matter,such as translational,vibrational,and rotational energies. It also includes the kinetic and potential energy of the nuclei and electrons within the individual molecules and the manner in which the molecules are linked together.
$E = E_{\text{translational}} + E_{\text{rotational}} + E_{\text{vibrational}} + E_{\text{electronic}} + E_{\text{nuclear}}$
Thus,we can say that internal energy is partly potential and partly kinetic.

(D) The formula for work done during expansion against a constant external pressure is $W = -P_{ext} \Delta V$.
Given: $P_{ext} = 3 \ atm$,$V_1 = 4 \ dm^3$,$V_2 = 6 \ dm^3$.
Since $1 \ dm^3 = 1 \ L$,$\Delta V = 6 \ L - 4 \ L = 2 \ L$.
$W = -3 \ atm \times 2 \ L = -6 \ L \ atm$.
Given $1 \ L \ atm = 101.32 \ J$,so $W = -6 \times 101.32 \ J = -607.92 \ J \approx -608 \ J$.

(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 $A_{(s)} + 3B_{(g)} \to 4C_{(s)} + D_{(l)}$,we calculate the change in the number of moles of gaseous species $(\Delta n_g)$:
$\Delta n_g = (n_g)_{\text{products}} - (n_g)_{\text{reactants}}$.
Since $C$ is solid and $D$ is liquid,they are not included in $\Delta n_g$.
$\Delta n_g = 0 - 3 = -3$.
Substituting this into the formula: $\Delta H = \Delta U + (-3)RT = \Delta U - 3RT$.

(D) 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 $(T_1 = T_2 = 300 \ K)$.
Therefore,the change in internal energy $(\Delta U)$ is given by $\Delta U = nC_v\Delta T$.
Since $\Delta T = 0$,the change in internal energy $\Delta U = 0$.

(A) For an isothermal reversible expansion of an ideal gas,the maximum work done is given by the formula: $W_{max} = -2.303 \ nRT \ \log(\frac{V_2}{V_1})$.
Given values are: $n = 0.75 \ mol$,$T = 27 \ ^{\circ}C = 300 \ K$,$V_1 = 15 \ L$,$V_2 = 25 \ L$,and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting these values: $W_{max} = -2.303 \times 0.75 \times 8.314 \times 300 \times \log(\frac{25}{15})$.
$W_{max} = -2.303 \times 0.75 \times 8.314 \times 300 \times \log(1.667)$.
$W_{max} = -2.303 \times 0.75 \times 8.314 \times 300 \times 0.2218 \approx -957.5 \ J$.
The magnitude of maximum work obtained is $957.5 \ J$.

(D) The molar enthalpy of fusion for ice at $0 \, ^{\circ}C$ is approximately $+6.01 \, kJ / mol$.
Therefore,the closest integer value is $6 \, kJ / mol$.

(A) In an exothermic reaction,heat is released to the surroundings.
This means the enthalpy of the products $(H_p)$ is less than the enthalpy of the reactants $(H_R)$,i.e.,$H_p < H_R$.
Therefore,the reactants have more energy than the products.
Thus,option $(A)$ is correct.

(A) The process of evaporation involves the conversion of liquid water into water vapour. This phase transition requires the absorption of energy in the form of latent heat. Therefore,it is an endothermic process.

(D) In an exothermic reaction,energy is released into the surroundings in the form of heat. Therefore,the reaction is accompanied by the evolution of heat,resulting in a negative change in enthalpy $(\Delta H < 0)$.

(B) An endothermic reaction is a chemical reaction that absorbs energy from its surroundings,typically in the form of heat.
Conversely,an exothermic reaction is a chemical reaction that releases energy into its surroundings,typically in the form of heat.

(A) The enthalpy change of a reaction is given by $\Delta H = H_P - H_R$,where $H_P$ is the enthalpy of products and $H_R$ is the enthalpy of reactants.
Given that the enthalpy of $B$ (product) is greater than that of $A$ (reactant),i.e.,$H_B > H_A$.
Therefore,$\Delta H = H_B - H_A > 0$.
$A$ reaction with a positive enthalpy change $(\Delta H > 0)$ is classified as an endothermic reaction.

(D) In an exothermic reaction,heat is released or evolved into the surroundings during the process.

(B) In a bomb calorimeter,the reaction is carried out at constant volume.
According to the first law of thermodynamics,the heat evolved at constant volume is equal to the change in internal energy,$\Delta E$ (or $\Delta U$).
Therefore,the heat of combustion measured in a bomb calorimeter is directly equal to $\Delta E$.
Given,$\Delta H_{combustion} = -870 \ kcal \ mol^{-1}$ (measured in a bomb calorimeter).
Thus,$\Delta E = -870 \ kcal \ mol^{-1}$.

(A) An exothermic reaction is a process in which net energy is released to the surroundings.
In the reaction $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}$,the total chemical potential energy of the reactants is higher than that of the product.
Since energy is conserved,the difference in energy is released as heat,making the reaction exothermic.