Basic concepts Questions in English

(D) The evaporation of water is a physical process represented as $H_2O(l) \rightarrow H_2O(g)$.
To convert liquid water into water vapor,energy must be supplied to overcome the intermolecular forces of attraction.
Since heat is absorbed from the surroundings during this process,it is an endothermic change.

(A) Work done by the system on the surroundings is given by the expression $w = -P_{ext} \Delta V$.
For the system to perform work on the surroundings (expansion),the volume must increase $(\Delta V > 0)$.
This expansion occurs when the internal pressure of the system is greater than the external pressure $(P_{sys} > P_{ext})$.
Therefore,energy is transferred as work from the system to the surroundings when the system pressure is greater than the atmospheric (external) pressure.

(C) An isolated system is defined as a system that cannot exchange either matter or energy with its surroundings.
Therefore,the correct option is $C$.

(C) bomb calorimeter measures the heat of combustion at constant volume.
By definition,the heat of reaction at constant volume is equal to the change in internal energy,$\Delta U$.
Therefore,for the combustion reaction of ethanol,$\Delta U = \Delta H_{combustion} = -670.48\, K\,cal\, mol^{-1}$.

(D) Intensive properties are those that do not depend on the amount of matter present in the system.
Temperature and Refractive Index are independent of the quantity of matter.
Enthalpy and Volume are extensive properties as they depend on the amount of substance.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ for a chemical reaction is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta n_g = N_P - N_r$ (change in the number of moles of gaseous species).
If $N_P > N_r$,then $\Delta n_g > 0$,which implies $\Delta H > \Delta U$.
If $N_P < N_r$,then $\Delta n_g < 0$,which implies $\Delta H < \Delta U$.
Therefore,the correct statement is $\Delta H > \Delta U$ when $N_P > N_r$.

(D) Intensive properties are those that do not depend on the amount of matter present in the system.
$1$. Enthalpy $(H)$ and Volume $(V)$ are extensive properties because they depend on the amount of substance.
$2$. Temperature $(T)$ and Refractive index $(n)$ are intensive properties because they are independent of the quantity of matter.
Therefore,the set (Temperature,Refractive index) consists of intensive properties.

(C) Initial volume $V_{initial} = 5 \, L$. Temperature $T = 27 + 273 = 300 \, K$.
Final volume $V_{final}$ is the volume occupied by $10 \, mol$ of gas at $1 \, atm$ pressure:
$V_{final} = \frac{nRT}{P} = \frac{10 \times 0.0821 \times 300}{1} = 246.3 \, L$.
Change in volume $\Delta V = V_{final} - V_{initial} = 246.3 - 5 = 241.3 \, L$.
Work done $W = -P_{ext} \Delta V = -1 \, atm \times 241.3 \, L = -241.3 \, L \cdot atm$.

(C) For an ideal gas,the enthalpy change $\Delta H$ is given by the relation $\Delta H = nC_p \Delta T$.
Since the process is isothermal,the temperature change $\Delta T = 0$.
Therefore,$\Delta H = nC_p(0) = 0 \, kJ$.
For an ideal gas,enthalpy is a function of temperature only,so for any isothermal process,$\Delta H = 0$.

(D) For an ideal gas,the number of moles $(n) = 1$; initial pressure $(P_1) = 10 \, atm$; final pressure $(P_2) = 1 \, atm$.
Constant absolute temperature $(T) = 300 \, K$; gas constant $(R) = 2 \, cal \, mol^{-1} \, K^{-1}$.
The work done $(W)$ for the reversible isothermal expansion of an ideal gas is given by:
$W = -2.303 \, nRT \, \log_{10} \left( \frac{P_1}{P_2} \right)$
$W = -2.303 \times 1 \times 2 \times 300 \times \log_{10} \left( \frac{10}{1} \right)$
$W = -2.303 \times 600 \times 1$
$W = -1381.8 \, cal$

(A) The work done by a system during expansion against an external pressure $P$ is given by the integral of the pressure with respect to the change in volume.
For a variable external pressure $P$,the work done $W$ is expressed as $W = -\int_{V_i}^{V_f} P_{ext} \, dV$.
Since the pressure $P$ is variable,the integral form is the correct representation of the work done.

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

(B) process is said to be reversible if it occurs infinitely slowly such that the system and surroundings are always in a state of equilibrium with each other at every stage of the process.
If the process is reversed,the system and surroundings return to their original states without leaving any net change in the universe.

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

(A) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 1 \times 10^5 \, N \, m^{-2}$.
Change in volume $\Delta V = V_f - V_i = (1 \times 10^{-2} - 1 \times 10^{-3}) \, m^3 = (10 \times 10^{-3} - 1 \times 10^{-3}) \, m^3 = 9 \times 10^{-3} \, m^3$.
Substituting the values: $W = -(1 \times 10^5) \times (9 \times 10^{-3}) = -900 \, J$.
Since the gas expands,work is done by the system,hence the value is negative.

(B) Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. Therefore,combustion is an exothermic process.

(D) Internal energy $(U)$ is a state function.
$A$ state function depends only on the initial and final states of the system,not on the path taken.
For any cyclic process,where the system returns to its initial state,the total change in the state function is always zero.
Since the system starts at state $A$,goes to state $B$,and returns to state $A$,the net change in internal energy $\Delta U_{net} = U_{final} - U_{initial} = U_A - U_A = 0$.

(D) Internal energy $(U)$ is the sum of all microscopic forms of energy in a system,including translational,rotational,vibrational,electronic,and nuclear energy.
It does not include macroscopic forms of energy such as kinetic energy due to the motion of the system as a whole or potential energy due to the gravitational pull on the system.
Therefore,the correct answer is $D$.

(C) The work done during expansion is given by the formula $W = -P_{ext} \Delta V$.
Since the gas expands into a vacuum,the external pressure $P_{ext} = 0$.
Therefore,$W = -0 \times \Delta V = 0 \ J$.
Thus,the work done is $0 \ J$.

(B) The heat exchanged in a chemical reaction at constant temperature and constant pressure is defined as the change in enthalpy $(\Delta H)$.
Mathematically, at constant pressure, the heat exchanged $(q_p)$ is equal to the change in enthalpy: $\Delta H = q_p$.

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g}RT$.
Depending on the value of $\Delta n_{g}$ (the change in the number of moles of gaseous species),$\Delta n_{g}RT$ can be positive,negative,or zero.
If $\Delta n_{g} > 0$,then $\Delta H > \Delta E$.
If $\Delta n_{g} < 0$,then $\Delta H < \Delta E$.
If $\Delta n_{g} = 0$,then $\Delta H = \Delta E$.
Therefore,$\Delta E$ may be lesser,greater,or equal to $\Delta H$.

(D) The work done in any reversible expansion process is given by the general expression $W = - \int_{1}^{2} P dV$.
For an adiabatic process,$q = 0$,and by the first law of thermodynamics,$\Delta U = W$.
Since $\Delta U = nC_v \Delta T$,the work done is $W = nC_v(T_2 - T_1)$.
However,the fundamental definition of reversible work for any gas expansion is $W = - \int_{1}^{2} P dV$.

(D) The work done during the expansion of a gas is given by the formula $W = -P_{ext} \Delta V$.
Since the gas expands into a vacuum,the external pressure $P_{ext}$ is $0$.
Therefore,the work done $W = -0 \times \Delta V = 0 \ J$.

(B) state function is a property whose value depends only on the state of the system and not on the path taken to reach that state.
$I$. $q + w = \Delta U$ (Internal energy change),which is a state function.
$II$. $q$ (Heat) is a path function.
$III$. $w$ (Work) is a path function.
$IV$. $H - TS = G$ (Gibbs free energy),which is a state function.
Therefore,$q$ and $w$ are not state functions.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$.
For $\Delta H = \Delta E$ to hold true,the change in the number of moles of gaseous species $(\Delta n_{g})$ must be $0$.
Calculating $\Delta n_{g}$ for option $B$: $H_{2_{(g)}} + Br_{2_{(g)}} \rightarrow 2HBr_{(g)}$.
$\Delta n_{g} = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 1) = 0$.
Since $\Delta n_{g} = 0$,it follows that $\Delta H = \Delta E$.

(D) The work done in an isothermal reversible expansion is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$
Using the base-$10$ logarithm: $W = -2.303 \times nRT \log(\frac{V_2}{V_1})$
Given values: $n = 6 \, mol$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$,$T = 27 + 273 = 300 \, K$,$V_1 = 1 \, L$,$V_2 = 10 \, L$.
Substituting the values: $W = -2.303 \times 6 \times 8.314 \times 300 \times \log(\frac{10}{1})$
$W = -2.303 \times 6 \times 8.314 \times 300 \times 1$
$W = -34464.8 \, J = -34.465 \, kJ$
The magnitude of the maximum work done is $34.465 \, kJ$.

(C) Initial volume,$V_{1} = 10 \, L$
Final volume,$V_{2} = \frac{nRT}{P} = \frac{10 \times 0.083 \times 300}{1} = 249 \, L$
Work done by the gas,$W = P_{ext} \Delta V = P_{ext} (V_{2} - V_{1})$
$W = 1 \, bar \times (249 \, L - 10 \, L) = 239 \, L \, bar$

(A) For an isothermal and reversible expansion of an ideal gas,the heat absorbed $(q)$ is given by the formula:
$q = -w = 2.303 \ nRT \ \log \left( \frac{V_2}{V_1} \right)$
Given:
$n = 0.75 \ mol$
$T = 27 + 273 = 300 \ K$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$V_1 = 15 \ L, V_2 = 25 \ L$
Substituting the values:
$q = 2.303 \times 0.75 \times 8.314 \times 300 \times \log \left( \frac{25}{15} \right)$
$q = 2.303 \times 0.75 \times 8.314 \times 300 \times \log \left( \frac{5}{3} \right)$
$q = 2.303 \times 0.75 \times 8.314 \times 300 \times (\log 5 - \log 3)$
$q = 2.303 \times 0.75 \times 8.314 \times 300 \times (0.70 - 0.48)$
$q = 2.303 \times 0.75 \times 8.314 \times 300 \times 0.22$
$q = 955.7 \ J$

(A) For an isothermal reversible process,the temperature $T$ is constant.
From the ideal gas law,$PV = nRT$,so $P = \frac{nRT}{V}$.
The work done $w$ is given by the integral:
$w = -\int_{V_1}^{V_2} P \, dV = -\int_{V_1}^{V_2} \frac{nRT}{V} \, dV$.
Since $n$,$R$,and $T$ are constants,we pull them out of the integral:
$w = -nRT \int_{V_1}^{V_2} \frac{1}{V} \, dV = -nRT [\ln V]_{V_1}^{V_2}$.
Therefore,$w = -nRT \ln \frac{V_2}{V_1}$.

(A) The enthalpy change $\Delta H$ is related to the internal energy change $\Delta U$ by the equation: $\Delta H = \Delta U + \Delta(PV)$.
Given $\Delta U = 35 \ \text{bar-L}$.
The change in the product of pressure and volume is $\Delta(PV) = P_2V_2 - P_1V_1$.
Substituting the given values: $P_1 = 2 \ \text{bar}, V_1 = 40 \ \text{L}, P_2 = 5 \ \text{bar}, V_2 = 15 \ \text{L}$.
$\Delta(PV) = (5 \ \text{bar} \times 15 \ \text{L}) - (2 \ \text{bar} \times 40 \ \text{L}) = 75 \ \text{bar-L} - 80 \ \text{bar-L} = -5 \ \text{bar-L}$.
Therefore, $\Delta H = 35 \ \text{bar-L} + (-5 \ \text{bar-L}) = 30 \ \text{bar-L}$.

(B) The work done by a gas during expansion is equal to the area under the $P-V$ curve.
From the given figure,the area under the curve for path $3$ is the largest,followed by path $2$,and the area under the curve for path $1$ is the smallest.
Therefore,the work done follows the order $w_3 > w_2 > w_1$,which is equivalent to $w_1 < w_2 < w_3$.

(A) In the given energy profile diagram for the reaction $R \to P$:
$a$ represents the potential energy of the reactant $R$ relative to the baseline.
$b$ represents the potential energy of the product $P$ relative to the baseline.
However,looking at the diagram,$a$ is the energy of $R$ and $b$ is the energy of $P$ relative to the same reference level.
The enthalpy change of the reaction,$\Delta H^o$,is defined as the difference between the potential energy of the products and the potential energy of the reactants.
$\Delta H^o = E_P - E_R$
From the diagram,$E_P = b$ and $E_R = a$.
Therefore,$\Delta H^o = b - a$.

(C) Intensive properties are those that do not depend on the quantity of matter present in the system.
Boiling point,$pH$,and $emf$ of a cell are intensive properties because they are independent of the amount of substance.
Entropy $(S)$ is an extensive property because it depends on the amount of matter present in the system.
Therefore,$I, III, \text{and } IV$ are intensive properties.

(A) Given: Number of moles $n = 1$.
Since the gas expands freely into a vacuum,the external pressure $P_{ext} = 0 \, atm$.
The formula for work done is $W = -P_{ext} \Delta V$.
Substituting the value of $P_{ext} = 0$,we get $W = -0 \times (V_2 - V_1) = 0$.
Therefore,the work done is zero.

(C) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta E + \Delta n_g RT$.
Rearranging this,we get $\Delta E - \Delta H = - \Delta n_g RT$.
For $(\Delta E - \Delta H)$ to be maximum,the value of $-\Delta n_g$ must be maximum,which means $\Delta n_g$ must be the most negative (minimum).
Calculating $\Delta n_g$ (moles of gaseous products - moles of gaseous reactants) for each reaction:
$A: \Delta n_g = (1 + 1) - 1 = 1$
$B: \Delta n_g = (1 + 1) - 0 = 2$
$C: \Delta n_g = 2 - (1 + 1) = 0$
$D: \Delta n_g = (0 + 1) - 0 = 1$
Since $\Delta E - \Delta H = - \Delta n_g RT$,the expression is maximum when $\Delta n_g$ is minimum. However,the question asks for the maximum value. Re-evaluating: if $\Delta n_g$ is negative,the value is positive. Among the given options,$C$ has the lowest $\Delta n_g$ $(0)$,making the value $0$. For $A, B, D$,$\Delta n_g$ is positive,making $(\Delta E - \Delta H)$ negative. Thus,the maximum value is $0$ for option $C$.

(D) The work done during expansion against constant external pressure is given by the formula: $W = -P_{ex} \Delta V$.
Given:
$P_{ex} = 1 \times 10^5 \,N/m^2$
$V_1 = 1 \times 10^{-3} \,m^3$
$V_2 = 1 \times 10^{-2} \,m^3 = 10 \times 10^{-3} \,m^3$
Change in volume:
$\Delta V = V_2 - V_1 = (10 \times 10^{-3} - 1 \times 10^{-3}) \,m^3 = 9 \times 10^{-3} \,m^3$
Calculating work:
$W = -(1 \times 10^5 \,N/m^2) \times (9 \times 10^{-3} \,m^3)$
$W = -9 \times 10^2 \,J = -900 \,J$

(A) 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 the reaction $CO_{(g)} + 1/2 O_{2_{(g)}} \to CO_{2_{(g)}}$,the change in the number of moles of gaseous species $(\Delta n_g)$ is calculated as:
$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$
$\Delta n_g = 1 - (1 + 0.5) = -0.5$.
Substituting this value into the equation:
$\Delta H = \Delta E + (-0.5) RT$
$\Delta H = \Delta E - 0.5 \, RT$.

(B) For an ideal gas,the enthalpy change $\Delta H$ is given by the relation $\Delta H = nC_p\Delta T$.
Since the process is isothermal,the change in temperature $\Delta T = 0$.
Therefore,$\Delta H = nC_p(0) = 0\, kJ$.

(D) Extensive properties are those properties of a system whose value depends on the quantity or size of matter present in the system. Examples include mass,volume,internal energy,enthalpy,and entropy $(S)$.
$1$. Molar entropy is an intensive property because it is defined per mole of substance.
$2$. Specific volume is an intensive property because it is defined as volume per unit mass.
$3$. Boiling point is an intensive property as it is independent of the amount of substance.
Since none of the given options are extensive properties,the correct answer is $D$.

(D) The work done during the expansion of a gas is given by the formula: $W = -P_{ext} \Delta V$.
Since the gas expands against a vacuum,the external pressure is $P_{ext} = 0 \, atm$.
Substituting the values into the equation: $W = -(0) \times (4 \, L - 1 \, L) = 0 \, J$.
Therefore,the work done is $0 \, J$.

(D) The standard enthalpy of formation,$\Delta_{f}H^{\circ}$,is defined as $0$ for an element in its most stable physical state at $25\,^{\circ}C$ $(298 \ K)$ and $1 \ bar$ pressure.
$1$. Carbon is most stable as $C(\text{graphite})$,not diamond.
$2$. Bromine exists as a liquid,$Br_{2(l)}$,at standard conditions.
$3$. Iodine exists as a solid,$I_{2(s)}$,at standard conditions.
Therefore,the set $C(\text{graphite}), Br_{2(l)}, I_{2(s)}$ represents elements in their standard states where $\Delta_{f}H^{\circ} = 0$.

(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 $\Delta H$ to be equal to $\Delta U$,the term $\Delta n_g$ must be $0$.
$\Delta n_g$ is the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants.
For option $B$: $2HI_{(g)} \to H_{2(g)} I_{2(g)}$,$\Delta n_g = (1 1) - 2 = 0$.
Since $\Delta n_g = 0$,$\Delta H = \Delta U 0$,which means $\Delta H = \Delta U$.

(B) For an ideal gas,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 $\Delta H = n C_p \Delta T = 0$,which means enthalpy remains constant.
For an expansion process,the final volume $V_f$ is greater than the initial volume $V_i$ $(V_f > V_i)$.
The change in entropy for an ideal gas is given by $\Delta S = n R \ln(V_f / V_i)$.
Since $V_f / V_i > 1$,$\ln(V_f / V_i) > 0$,therefore $\Delta S > 0$,which means entropy increases.

(B) For a reversible isothermal expansion,the work done by the gas is given by $w = -nRT \ln(\frac{V_f}{V_i})$.
Taking the magnitude,we have $|w| = nRT \ln V_f - nRT \ln V_i$.
This equation is in the form of a straight line $y = mx + c$,where $y = |w|$,$x = ln V_f$,slope $m = nRT$,and intercept $c = -nRT \ln V_i$.
Since $T_2 > T_1$,the slope for $T_2$ $(nRT_2)$ is greater than the slope for $T_1$ $(nRT_1)$.
Also,for a given expansion,the intercept $-nRT \ln V_i$ will be more negative for higher temperatures if we assume the same initial state or consistent expansion parameters,leading to the graph where the line for $T_2$ is steeper and has a lower intercept than the line for $T_1$.

(A) Path functions are properties whose values depend on the path taken to reach a state,not just the initial and final states.
$q$ (heat) and $w$ (work) are path functions.
$q + w$ represents the change in internal energy $(\Delta U)$,which is a state function.
$H - TS$ represents the Gibbs free energy $(G)$,which is a state function.
Therefore,only $q$ and $w$ are path functions.

(A) The work done during expansion against a constant external pressure is given by the formula: $W = -P_{ext} \cdot \Delta V$.
Given: $P_{ext} = 1 \, bar$, $V_1 = 1 \, L$, $V_2 = 10 \, L$.
$\Delta V = V_2 - V_1 = 10 \, L - 1 \, L = 9 \, L$.
$W = -1 \, bar \cdot 9 \, L = -9 \, bar \cdot L$.
Since $1 \, bar \cdot L = 100 \, J$, then $W = -9 \times 100 \, J = -900 \, J$.
Converting to $kJ$: $W = -900 \, J / 1000 = -0.9 \, kJ$.

(B) For a reversible isothermal compression of an ideal gas,the work done is given by the formula: $W = -2.303 \, nRT \log \frac{P_1}{P_2}$
Given: $n = 1 \, mol$,$T = 27 + 273 = 300 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$P_1 = 1 \, bar$,$P_2 = 4 \, bar$.
Substituting the values: $W = -2.303 \times 1 \times 8.314 \times 300 \times \log(\frac{1}{4})$
$W = -2.303 \times 8.314 \times 300 \times (-0.602)$
$W = +3.458 \, kJ$.

(D) Extensive properties are those that depend on the amount of matter present in the system (e.g.,mass,volume,internal energy).
Intensive properties are those that are independent of the amount of matter present in the system.
Temperature,pressure,and viscosity are all independent of the amount of matter in the system,meaning they are intensive properties.
Therefore,all of the given options are not extensive properties.

(B) The work done by a gas during an expansion process is equal to the area under the $PV$ curve.
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.
- The area under path $2$ is intermediate.
- The area under path $1$ is the smallest.
Therefore,the work done follows the order: $w_1 < w_2 < w_3$.