Basic concepts Questions in English

(N/A) $1$. Open System: $A$ system that can exchange both energy and matter with its surroundings. Example: Boiling water in an open beaker.
$2$. Closed System: $A$ system that can exchange energy but not matter with its surroundings. Example: Boiling water in a sealed container.
$3$. Isolated System: $A$ system that cannot exchange either energy or matter with its surroundings. Example: Hot water in a perfectly insulated thermos flask.

(A) An open system is one that can exchange both matter and energy with its surroundings.
In an open vessel,a saturated solution of glucose can exchange both water vapor (matter) and heat (energy) with the surroundings.
Therefore,it is an open system.

(N/A) $1$. Exothermic Reaction: $A$ chemical reaction in which energy is released into the surroundings,usually in the form of heat or light. In these reactions,the total energy of the products is less than the total energy of the reactants. Example: Combustion of methane,$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{Heat}$.
$2$. Endothermic Reaction: $A$ chemical reaction in which energy is absorbed from the surroundings. In these reactions,the total energy of the products is greater than the total energy of the reactants. Example: Photosynthesis,$6CO_2 + 6H_2O + \text{Energy} \rightarrow C_6H_{12}O_6 + 6O_2$.

(N/A) At constant volume,the heat capacity is denoted by $C_v$ and at constant pressure,it is denoted by $C_p$.
For a process at constant volume,the heat exchanged is $q_v = \Delta U = C_v \Delta T$.
For a process at constant pressure,the heat exchanged is $q_p = \Delta H = C_p \Delta T$.
For $1 \text{ mole}$ of an ideal gas,the enthalpy change is given by $\Delta H = \Delta U + \Delta (pV)$.
Since $pV = RT$ for $1 \text{ mole}$ of an ideal gas,we have $\Delta H = \Delta U + \Delta (RT)$.
Assuming $R$ is constant,$\Delta H = \Delta U + R \Delta T$.
Substituting the expressions for $\Delta H$ and $\Delta U$ in terms of heat capacities:
$C_p \Delta T = C_v \Delta T + R \Delta T$.
Dividing both sides by $\Delta T$,we get $C_p = C_v + R$ or $C_p - C_v = R$.

(A) The formula for work done by a system during expansion against a constant external pressure is $w = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 10 \ bar$,$V_1 = 100 \ L$,and $V_2 = 120 \ L$.
Change in volume $\Delta V = V_2 - V_1 = 120 \ L - 100 \ L = 20 \ L$.
Substituting the values: $w = -10 \ bar \times 20 \ L = -200 \ L \ bar$.
Given the conversion factor $1 \ L \ bar = 24.21 \ cal$,we convert the work to calories:
$w = -200 \times 24.21 \ cal = -4842 \ cal$.
The negative sign indicates that work is done by the system.

(N/A) We can measure energy changes associated with chemical or physical processes by an experimental technique called calorimetry.
In calorimetry,the process is carried out in a vessel called a calorimeter,which is immersed in a known volume of a liquid.
Knowing the heat capacity of the liquid in which the calorimeter is immersed and the heat capacity of the calorimeter,it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions:
$(i)$ at constant volume,$q_{v}$
$(ii)$ at constant pressure,$q_{p}$
$(a)$ $\Delta U$ measurements:
For chemical reactions,heat absorbed at constant volume is measured in a bomb calorimeter. Here,a steel vessel is immersed in a water bath. The whole device is called a calorimeter. The steel vessel is immersed in a water bath to ensure that no heat is lost to the surroundings.
$(b)$ $\Delta H$ measurements:
Heat absorbed at constant pressure is measured using a calorimeter (often a coffee-cup calorimeter). Since $\Delta H = q_{p}$,the heat change measured at constant pressure directly gives the enthalpy change of the reaction.

(N/A) The enthalpy of a reaction depends on the conditions under which the reaction is carried out. Therefore,it is necessary to specify standard conditions.
The standard enthalpy of reaction is defined as the enthalpy change for a reaction when all the participating substances are in their standard states.
The standard state of a substance at a specified temperature is its pure form at $1 \ bar$ pressure.
For example,the standard state of liquid ethanol at $298 \ K$ is pure liquid ethanol at $1 \ bar$; the standard state of solid iron at $500 \ K$ is pure iron at $1 \ bar$. Usually,thermodynamic data are reported at $298 \ K$.
Standard conditions are denoted by adding the superscript $\theta$ to the symbol $\Delta H$,represented as $\Delta H^{\theta}$.

(N/A) Standard enthalpy of fusion $(\Delta_{fus} H)$: The enthalpy change that accompanies the melting of one mole of a solid substance in its standard state is called the standard enthalpy of fusion. Melting is an endothermic process,so $\Delta_{fus} H$ is always positive.
$H_2O_{(s)} \rightarrow H_2O_{(l)}$; $\Delta_{fus} H = 6.00 \ kJ \ mol^{-1}$
Standard enthalpy of vaporization $(\Delta_{vap} H)$: The amount of heat required to vaporize one mole of a liquid at a constant temperature and under standard pressure $(1 \ bar)$ is called its standard enthalpy of vaporization.
$H_2O_{(l)} \rightarrow H_2O_{(g)}$; $\Delta_{vap} H = +40.79 \ kJ \ mol^{-1}$
Standard enthalpy of sublimation $(\Delta_{sub} H)$: The change in enthalpy when one mole of a solid substance sublimes directly into its vapour at a constant temperature and under standard pressure $(1 \ bar)$.
$CO_{2_{(s)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{sub} H^o = 25.2 \ kJ \ mol^{-1}$
The magnitude of these enthalpy changes depends on the strength of the intermolecular interactions in the substance.

Substance	$T_f / K$	$\Delta_{fus} H / (kJ \ mol^{-1})$	$T_b / K$	$\Delta_{vap} H / (kJ \ mol^{-1})$
$N_2$	$63.15$	$0.72$	$77.35$	$5.59$
$NH_3$	$195.40$	$5.65$	$239.73$	$23.35$
$HCl$	$159.0$	$1.992$	$188.0$	$16.15$
$CO$	$68.0$	$6.836$	$82.0$	$6.04$
$CH_3COCH_3$	$177.8$	$5.72$	$329.4$	$29.1$
$CCl_4$	$250.16$	$2.5$	$349.69$	$30.0$
$H_2O$	$273.15$	$6.01$	$373.15$	$40.79$
$NaCl$	$1081.0$	$28.8$	$1665.0$	$170.0$
$C_6H_6$	$278.65$	$9.83$	$353.25$	$30.8$

Enthalpy of dilution is the enthalpy change observed when a specified amount of solvent is added to a solution at a constant temperature and pressure.
For example,the enthalpy change for dissolving $1 \ mol$ of gaseous hydrogen chloride in $10 \ mol$ of water is represented by:
$HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O ; \Delta H = -69.01 \ kJ/mol$
Consider the following enthalpy changes for different amounts of water:
$(S-1) \ HCl_{(g)} + 25 \ H_2O_{(l)} \rightarrow HCl \cdot 25 \ H_2O ; \Delta H = -72.03 \ kJ/mol$
$(S-2) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O ; \Delta H = -72.79 \ kJ/mol$
$(S-3) \ HCl_{(g)} + \infty \ H_2O_{(l)} \rightarrow HCl \cdot \infty \ H_2O ; \Delta H = -74.85 \ kJ/mol$
As more solvent is added,the enthalpy of solution approaches a limiting value at infinite dilution,as shown in $(S-3)$.
To find the enthalpy of dilution between two concentrations,subtract $(S-1)$ from $(S-2)$:
$(HCl \cdot 25 \ H_2O) + 15 \ H_2O \rightarrow HCl \cdot 40 \ H_2O ; \Delta H = [-72.79 - (-72.03)] = -0.76 \ kJ/mol$
This value,$-0.76 \ kJ/mol$,is the enthalpy of dilution. It depends on the initial concentration of the solution and the amount of solvent added.

(N/A) Yes,the temperature of the system and the surroundings are the same when they are in thermal equilibrium.
Thermal equilibrium is defined as a state where two physical systems are in contact through a diathermal wall,allowing the transfer of energy as heat,but no net transfer of energy occurs between them.
When two systems are in thermal equilibrium,there is no change in their states,which implies that their temperatures are equal.
This concept is governed by the $Zeroth \ Law \ of \ Thermodynamics$.

(0) Enthalpy $(H)$ is a state function,which means its value depends only on the initial and final states of the system.
In a cyclic process,the system returns to its initial state after completing the cycle.
Therefore,the change in enthalpy for the entire cycle is zero,i.e.,$\Delta H_{\text{cycle}} = 0$.

(N/A) State functions are properties that depend only on the initial and final states of the system,not on the path taken to reach that state. Examples include: enthalpy $(H)$,entropy $(S)$,temperature $(T)$,and free energy $(G)$.
Path functions are properties that depend on the specific path or process by which the system changes from one state to another. Examples include: heat $(q)$ and work $(w)$.

(C) For an isolated system,there is no transfer of energy as heat,i.e.,$q = 0$.
There is no transfer of energy as work,i.e.,$W = 0$.
According to the first law of thermodynamics:
$\Delta U = q + W$
$\Delta U = 0 + 0 = 0$

(N/A) The two conditions under which heat becomes independent of path are:
$(i)$ When volume remains constant $(q_V)$
$(ii)$ When pressure remains constant $(q_p)$
Explanation:
$(i)$ At constant volume: By the first law of thermodynamics,$\Delta U = q + W$. Since $W = -P_{ext} \Delta V$ and $\Delta V = 0$,the work done is $0$. Therefore,$q_V = \Delta U$. Since internal energy $(\Delta U)$ is a state function,$q_V$ is also a state function.
$(ii)$ At constant pressure: The heat absorbed is given by $q_p = \Delta U + P \Delta V$. By definition,the change in enthalpy is $\Delta H = \Delta U + P \Delta V$. Therefore,$q_p = \Delta H$. Since enthalpy $(\Delta H)$ is a state function,$q_p$ is also a state function.

(A) Work done of a gas in vacuum is given by the formula $W = -p_{ext} \Delta V$.
Since the expansion occurs in vacuum,the external pressure $p_{ext} = 0$.
Therefore,$W = -0 \times (5 \ L - 1 \ L) = 0 \ J$.
For an ideal gas,the internal energy $U$ is a function of temperature only $(U = f(T))$.
Since the process is isothermal,the temperature remains constant $(\Delta T = 0)$,which implies that the change in internal energy $\Delta U = 0$.

(A) For an ideal gas,the enthalpy is defined as $H = U + PV$.
Since $PV = nRT$ for $n$ moles of an ideal gas,we have $H = U + nRT$.
Differentiating with respect to temperature $T$ at constant pressure,we get $\frac{dH}{dT} = \frac{dU}{dT} + nR$.
By definition,$C_p = \frac{dH}{dT}$ and $C_v = \frac{dU}{dT}$.
Therefore,$C_p - C_v = nR$.
Given $n = 10$,the difference is $10R$.

(N/A) The change in volume is given by $\Delta V = (V_f - V_i)$.
If $W$ is the work done on the system by the movement of the piston,then the work done is defined as:
$W = -p_{ext} \times \Delta V$
Substituting $\Delta V = (V_f - V_i)$,we get:
$W = -p_{ext} \times (V_f - V_i) = p_{ext} \times (V_i - V_f)$
This can be represented on a $p-V$ graph as shown in the figure. The work done is equal to the shaded area under the pressure-volume curve.
In the case of compression,the system is compressed from initial volume $V_i$ to final volume $V_f$,where $V_f < V_i$. Thus,$\Delta V$ is negative,making the work done $W$ positive,which is consistent with the sign convention that work done on the system is positive.

$A$ process or change is said to be reversible if it is carried out in such a way that the process could,at any moment,be reversed by an infinitesimal change.
When pressure changes in infinite steps (reversible conditions) during compression from an initial volume $V_i$ to a final volume $V_f$,the work done can be calculated by integrating the pressure with respect to volume:
$w = -\int_{V_i}^{V_f} p_{ext} \, dV$
In a $pV$-plot,this work done on the gas is represented by the area under the curve between $V_i$ and $V_f$.

(N/A) The given diagram represents an isothermal reversible expansion of an ideal gas from pressure $P_1 = 2.0 \ bar$ to $P_2 = 1.0 \ bar$ at $T = 298 \ K$.
For an isothermal reversible process,the work done $W$ is given by the formula:
$W = -2.303 \ nRT \log_{10} \left( \frac{P_1}{P_2} \right)$
Given:
$n = 1.0 \ mol$
$R = 8.314 \ J \ K^{-1} \ mol^{-1}$
$T = 298 \ K$
$P_1 = 2.0 \ bar$
$P_2 = 1.0 \ bar$
Substituting the values:
$W = -2.303 \times 1.0 \times 8.314 \times 298 \times \log_{10} \left( \frac{2.0}{1.0} \right)$
$W = -2.303 \times 8.314 \times 298 \times 0.3010$
$W \approx -1717.46 \ J \approx -1.717 \ kJ$

(N/A) In the first case,as the expansion is against constant external pressure,
$W = -p_{ext} \times (V_2 - V_1) = -2 \ bar \times (50 - 10) \ L = -80 \ L \ bar$
Since $1 \ L \ bar = 100 \ J$,
$W = -80 \times 100 \ J = -8000 \ J = -8 \ kJ$
If the given expansion was carried out reversibly,the internal pressure of the gas would be infinitesimally greater than the external pressure at every stage,resulting in maximum work done. Therefore,the work done will be higher than in the irreversible case.

(N/A) Extensive properties: Properties whose values depend on the quantity or size of matter present in the system are known as extensive properties.
Examples: $Mass$,$internal \ energy$,$heat \ capacity$.
Intensive properties: Properties that do not depend on the quantity or size of matter present are known as intensive properties.
Examples: $Pressure$,$molar \ heat \ capacity$,$density$,$mole \ fraction$,$specific \ heat$,$temperature$,and $molarity$.
Note: The ratio of two extensive properties is always an intensive property.
$\frac{\text{Extensive}}{\text{Extensive}} = \text{Intensive}$
For example,$mole \ fraction = \frac{\text{Moles of component}}{\text{Total moles}} = \frac{\text{Extensive}}{\text{Extensive}}$ and $molarity = \frac{\text{Moles}}{\text{Volume}} = \frac{\text{Extensive}}{\text{Extensive}}$. Thus,they are intensive properties.

(N/A) $(i)$ Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from $(P_i, V_i)$ to $(P_f, V_f)$ is represented by the area under the curve between $V_i$ and $V_f$.
$(ii)$ Work done against constant external pressure $P_f$ is represented by the rectangular area under the line at $P_f$ between $V_i$ and $V_f$.
$(iii)$ Comparing the two,the area under the reversible curve is greater than the area under the constant pressure line,therefore,reversible work is greater than the work done against constant external pressure $P_f$.

(A) In thermodynamics,a $System$ is defined as that part of the universe which is under investigation or chosen for study.
Everything else in the universe,excluding the system,is called the $Surroundings$.
The system and surroundings together constitute the $Universe$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
For the heat absorbed $(q)$ to be completely converted into work $(-w)$,the change in internal energy $(\Delta U)$ must be zero.
This condition,$\Delta U = 0$,is characteristic of an isothermal process for an ideal gas.

(B) Free expansion is defined as the expansion of a gas into a vacuum.
Since the external pressure $(P_{ext})$ is $0$,the work done $(w)$ by the system is calculated as $w = -P_{ext} \Delta V$.
Because $P_{ext} = 0$,the work done is $w = 0$.
Therefore,free expansion refers to expansion against a vacuum.

(A) Extensive properties are those properties whose values depend on the quantity or size of matter present in the system.
Examples include mass,volume,internal energy,enthalpy,and entropy.
In contrast,intensive properties are independent of the amount of matter present.

(B) Calorimetry is the experimental technique used to measure the energy changes associated with chemical and physical processes.

(B) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{(g)}RT$.
For $\Delta H = \Delta U$ to hold true,the term $\Delta n_{(g)}RT$ must be equal to $0$.
Since $R$ (gas constant) and $T$ (temperature) are non-zero,this implies $\Delta n_{(g)} = 0$.

(C) The standard state of a substance at a specified temperature is defined as its pure form at $1 \, \text{bar}$ pressure.
By convention,$298 \, K$ is taken as the standard temperature for thermodynamic calculations.
Therefore,the standard state corresponds to $1 \, \text{bar}$ pressure and $298 \, K$ temperature.

(N/A) The molar enthalpy of vaporization,also known as the standard enthalpy of vaporization,is the amount of heat required to vaporize one mole of a liquid at constant temperature and under constant pressure.

(N/A) The relationship between the boiling point $(T_b)$ and the enthalpy of vaporization $(\Delta_{vap}H)$ of a solvent is described by Trouton's rule. For many liquids,the entropy of vaporization $(\Delta_{vap}S)$ is approximately constant at the boiling point. The relation is given by: $\Delta_{vap}S = \frac{\Delta_{vap}H}{T_b} \approx 88 \ J \ K^{-1} \ mol^{-1}$.

(N/A) Extensive properties are those properties whose values depend on the quantity or size of matter present in the system. Examples include mass,volume,internal energy,and enthalpy.

(N/A) Calorimetry is the experimental technique used to measure the energy changes associated with chemical and physical processes.

(B) The relationship between enthalpy change and internal energy change is given by the equation: $\Delta H = \Delta U + \Delta n_{(g)}RT$.
For $\Delta H = \Delta U$ to hold true,the term $\Delta n_{(g)}RT$ must be equal to $0$.
Since $R$ (gas constant) and $T$ (temperature) are non-zero,it implies that $\Delta n_{(g)} = 0$.

(N/A) The standard state is defined as a pressure of $1 \, \text{bar}$ (or $1 \, \text{atm}$) and a temperature of $298 \, K$.

(N/A) The molar enthalpy of vaporization is the amount of heat required to vaporize $1 \ mol$ of a liquid at a constant temperature and under constant pressure.

(C) The enthalpy of dilution is the heat change associated with the process of diluting a solution.
It depends on the initial concentration of the solution and the amount of solvent added to it.

(C) The $Zeroth \ law$ of thermodynamics states that if two systems are each in thermal equilibrium with a third system,they are in thermal equilibrium with each other. This law provides the basis for the measurement of temperature and is therefore considered the definition of temperature.

(C) Intensive properties are those that do not depend on the amount of matter present in the system.
Examples include:
$1$. Melting point
$2$. Boiling point
$3$. Density
$4$. Refractive index
$5$. Molar heat capacity
$6$. Pressure
$7$. Temperature

(A) $(i)$ $A$ system that exchanges energy but not matter with its surroundings is known as a $Closed \ system$.
$(ii)$ The sum of internal energy $(U)$ and the product of pressure $(P)$ and volume $(V)$ is defined as $Enthalpy$ $(H = U + PV)$.
$(iii)$ The heat content change or energy change during a reaction at constant pressure is known as $Enthalpy \ change$ $(\Delta H)$.

(A) $(i)$ Zeroth law of thermodynamics.
$(ii)$ Conservation of energy (First law).
$(iii)$ Zeroth law of thermodynamics.

(N/A) $(i)$ $1 \ \text{calorie} = 4.184 \ \text{joules}$
$(ii)$ The work done by the system is given by the equation $W = -P_{ext} \Delta V$.

(N/A) $(i)$ Internal energy change (or $\Delta U$)
$(ii)$ Adiabatic
$(iii)$ Gibbs-Helmholtz

(N/A) The energy that arises due to the motion of atoms or molecules in a body is called $thermal \ energy$. It is a measure of the average kinetic energy of particles,which increases with an increase in temperature.

(N/A) The enthalpy of dilution of a solution depends on the initial concentration of the solution and the amount of solvent added to it.

(C) The $Zeroth$ $Law$ of thermodynamics provides the definition of temperature. It states that if two systems are each in thermal equilibrium with a third system,they are in thermal equilibrium with each other.

(A) $(i)$ $A$ system that allows energy exchange but not matter exchange is a $\text{Closed system}$.
$(ii)$ The sum of internal energy $(U)$ and the product of pressure $(P)$ and volume $(V)$ is defined as $\text{Enthalpy}$ $(H = U + PV)$.
$(iii)$ The energy change at constant pressure is defined as $\text{Enthalpy change}$ $(\Delta H)$.
$(iv)$ The law describing thermal equilibrium is the $\text{Zeroth law of thermodynamics}$.

(N/A) $(i)$ Endothermic (or non-spontaneous at standard conditions)
$(ii)$ $4.184 \ J$
$(iii)$ $W = -P_{ext} \Delta V$
$(iv)$ Change in internal energy $(\Delta U)$

(N/A) $(i)$ Adiabatic process
$(ii)$ Gibbs-Helmholtz

(A) thermal power plant operates on the principle of energy conversion:
$1$. Fuel (coal,oil,or gas) is burned in a boiler to produce heat.
$2$. This heat is used to convert water into high-pressure steam.
$3$. The steam is directed onto the blades of a turbine,causing it to rotate (kinetic energy).
$4$. The rotating turbine is connected to a generator,which converts the mechanical energy into electrical energy through electromagnetic induction.