Basic concepts Questions in English

(D) The expansion of an ideal gas into a vacuum is known as free expansion.
In free expansion,the external pressure $(p_{\text{ext}})$ is $0 \ bar$.
The formula for work done is $W = -p_{\text{ext}} \Delta V$.
Since $p_{\text{ext}} = 0$,the work done $W = 0$.

(A) For an ideal gas:
$1$. Internal energy $U$ and enthalpy $H$ are functions of temperature only,i.e.,$U = f(T)$ and $H = f(T)$. Thus,statement $(a)$ is true.
$2$. The compressibility factor $Z$ for an ideal gas is defined as $Z = \frac{PV}{nRT} = 1$. Thus,statement $(b)$ is false.
$3$. For an ideal gas,the molar heat capacity relation is $C_{P,m} - C_{V,m} = R$. Thus,statement $(c)$ is true.
$4$. The relation $dU = C_V dT$ holds for an ideal gas for any process,whether reversible or irreversible. Thus,statement $(d)$ is true.
Therefore,statements $(a), (c),$ and $(d)$ are correct.

(C) The change in enthalpy is defined as $\Delta H = \Delta U + \Delta(PV)$.
Given $\Delta U = 30.0 \, L \, atm$.
Initial state: $P_1 = 2.0 \, atm$,$V_1 = 3.0 \, L$.
Final state: $P_2 = 4.0 \, atm$,$V_2 = 5.0 \, L$.
$\Delta(PV) = P_2 V_2 - P_1 V_1 = (4.0 \times 5.0) - (2.0 \times 3.0) = 20.0 - 6.0 = 14.0 \, L \, atm$.
Therefore,$\Delta H = 30.0 + 14.0 = 44.0 \, L \, atm$.

(B) For one mole of an ideal gas,the relationship between molar heat capacity at constant pressure $(C_{P})$ and molar heat capacity at constant volume $(C_{V})$ is given by Mayer's relation:
$C_{P} - C_{V} = R$
where $R$ is the universal gas constant.

(D) For an isothermal reversible expansion,the maximum work $(W_{max})$ is given by the formula:
$W_{max} = -nRT \ln \frac{V_2}{V_1} = -2.303 nRT \log \frac{V_2}{V_1}$
Given:
$n = 5 \ mol$,$T = 300 \ K$,$V_1 = 10 \ L$,$V_2 = 20 \ L$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$,$\log 2 = 0.3010$
Substituting the values:
$W_{max} = -2.303 \times 5 \times 8.3 \times 300 \times \log \frac{20}{10}$
$W_{max} = -2.303 \times 5 \times 8.3 \times 300 \times 0.3010$
$W_{max} = -8630.38 \ J$
The magnitude of the work is $|W_{max}| = 8630.38 \ J$.
Rounding to the nearest integer,we get $8630 \ J$.

(A) The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_{g} RT$.
For the reaction $H_{2}F_{2(g)} \rightarrow H_{2(g)} + F_{2(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = (1 + 1) - 1 = 1$.
The temperature is $T = 27 + 273 = 300 \ K$.
Given $\Delta U = -59.6 \ kJ \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
Substituting the values: $\Delta H = -59.6 + (1 \times 8.314 \times 10^{-3} \times 300) = -59.6 + 2.4942 = -57.1058 \ kJ \ mol^{-1}$.
The nearest integer value is $-57 \ kJ \ mol^{-1}$.

(B) State variables are properties whose values depend only on the state of the system and not on the path taken to reach that state.
In the given list:
$1$. Internal energy $(U)$ is a state function.
$2$. Volume $(V)$ is a state function.
$3$. Enthalpy $(H)$ is a state function.
$4$. Heat $(q)$ is a path function,not a state function.
Therefore,there are $3$ state variables: Internal energy,Volume,and Enthalpy.

(C) For an ideal gas,the internal energy $ \Delta U $ and enthalpy $ \Delta H $ are functions of temperature only.
For an isothermal process,the change in temperature $ \Delta T = 0 $.
Since $ \Delta U = n C_V \Delta T $,if $ \Delta T = 0 $,then $ \Delta U = 0 $.
Similarly,since $ \Delta H = \Delta U + n R \Delta T $,if $ \Delta T = 0 $ and $ \Delta U = 0 $,then $ \Delta H = 0 $.
Therefore,for an isothermal reversible expansion of an ideal gas,both $ \Delta H = 0 $ and $ \Delta U = 0 $.

(D) Intensive variables are those properties of a system that are independent of the amount or size of matter present in the system.
Density $(\rho)$,temperature $(T)$,and pressure $(p)$ are intensive properties because they do not change with the amount of substance.
Enthalpy $(H)$,heat capacity $(C_p)$,and volume $(V)$ are extensive properties because they depend on the quantity of matter present.
Therefore,the set of intensive variables is $(\rho, T, p)$.
Thus,the correct option is $D$.

(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 $(\Delta T = 0)$.
Therefore,the change in internal energy $\Delta U = 0 \ J$.

(A) Intensive properties are those that do not depend on the quantity or size of matter present in the system.
$1$. Volume: Extensive (depends on amount).
$2$. Molar heat capacity: Intensive (defined per mole).
$3$. Molarity: Intensive (concentration is independent of amount).
$4$. $E^{\theta}_{cell}$: Intensive (potential is independent of amount).
$5$. Gibbs free energy change $(\Delta G)$: Extensive (depends on amount).
$6$. Molar mass: Intensive (property of the substance itself).
$7$. Mole: Extensive (measure of amount).
Therefore,the intensive properties are: Molar heat capacity,Molarity,$E^{\theta}_{cell}$,and Molar mass.
The total count is $4$.

(A) The combustion of methane $(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O)$ is an exothermic process.
In System $A$,the temperature rises because the system is adiabatic,meaning no heat can escape to the surroundings.
In System $B$,the temperature remains constant because the system is diathermic,allowing heat to be exchanged with the surroundings to maintain thermal equilibrium.
Therefore,System $A$ is an adiabatic system and System $B$ is a diathermic system.

(C) The enthalpy change $(\Delta H)$ is defined as the change in internal energy $(\Delta U)$ plus the work done due to volume change at constant pressure.
For a chemical reaction involving gases,the relation is given by:
$\Delta H = \Delta U + \Delta n_{g} R T$
Where:
$\Delta H$ = Change in enthalpy
$\Delta U$ = Change in internal energy
$\Delta n_{g}$ = Change in the number of moles of gaseous products and reactants
$R$ = Universal gas constant
$T$ = Temperature in Kelvin

(B) For an isothermal reversible expansion,the work done is given by the formula:
$w = -2.303 \ nRT \log \left( \frac{V_2}{V_1} \right)$
Substituting the given values:
$n = 5 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$V_1 = 10 \ L$,$V_2 = 100 \ L$
$w = -2.303 \times 5 \times 8.314 \times 300 \times \log \left( \frac{100}{10} \right)$
$w = -2.303 \times 5 \times 8.314 \times 300 \times \log(10)$
Since $\log(10) = 1$:
$w = -2.303 \times 5 \times 8.314 \times 300 = -28720.713 \ J$
Given $w = -x \ J$,we have $-x = -28720.713$,so $x \approx 28721$.

(C) Isothermal process $\Rightarrow$ Temperature is constant throughout the process.
$(B)$ Isochoric process $\Rightarrow$ Volume is constant throughout the process.
$(C)$ Isobaric process $\Rightarrow$ Pressure is constant throughout the process.
$(D)$ Adiabatic process $\Rightarrow$ No exchange of heat $(q)$ between the system and surroundings.
Therefore,the correct matching is: $A-II, B-III, C-IV, D-I$.

(A) For a reversible isothermal expansion,the work done is given by the formula:
$W = -2.303 \ nRT \log \frac{P_i}{P_f}$
Given:
$n = 1 \ mol$
$R = 2.0 \ cal \ K^{-1} \ mol^{-1}$
$T = 25^{\circ}C = 298 \ K$
$P_i = 20 \ atm$
$P_f = 10 \ atm$
Substituting the values:
$W = -2.303 \times 1 \times 2.0 \times 298 \times \log \frac{20}{10}$
$W = -2.303 \times 2.0 \times 298 \times \log 2$
$W = -2.303 \times 596 \times 0.3010$
$W \approx -413.14 \ calories$

(B) The correct option is $B$.
Explanation:
$1$. $A$ 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.
$2$. Internal energy $(U)$ and molar enthalpy $(H_m)$ are state functions because they depend only on the state variables of the system.
$3$. Work $(w)$ and heat $(q)$ are path functions,meaning their values depend on the specific process or path taken to change the state of the system.
$4$. Therefore,both $A$ (Internal energy) and $D$ (Molar enthalpy) are state functions.

(B) By definition,the standard molar enthalpy of formation $(\Delta_fH^circ)$ of an element in its most stable state at $298 \ K$ and $1 \ \text{bar}$ pressure is $ZERO$.
$Cl_{2(g)}$ is the most stable form of chlorine at $298 \ K$.
$Br_{2(l)}$ is the most stable form of bromine at $298 \ K$,so $Br_{2(g)}$ does not have a zero enthalpy of formation.
$H_2O_{(g)}$ and $CH_{4(g)}$ are compounds,not elements,so their standard enthalpy of formation is not zero.

(C) The van der Waals equation for $1$ mole of gas is given by $P = \frac{RT}{V - b} - \frac{a}{V^2}$.
In thermodynamics,the work done is defined as $w = -\int P_{ext} dV$.
If the process is reversible,the external pressure $P_{ext}$ is equal to the internal pressure $P$ of the system at every stage,i.e.,$P_{ext} = P$.
Substituting the van der Waals pressure into the work equation gives $w = -\int \left(\frac{RT}{V - b} - \frac{a}{V^2}\right) dV$.
This expression is valid for any reversible process involving a van der Waals gas,regardless of whether it is isothermal or adiabatic. Thus,both $A$ and $B$ (in the context of reversibility) are satisfied.

(D) Internal energy $(U)$ is a state function,which means its value depends only on the state of the system and not on the path taken.
In any cyclic process,the system returns to its initial state.
Therefore,the change in internal energy $(\Delta U)$ for a complete cycle is always zero.
$\Delta U_{\text{cycle}} = U_{\text{final}} - U_{\text{initial}} = 0$.
Since all three cases represent cyclic processes starting and ending at point $A$,the change in internal energy for each case is equal to zero.
Thus,$\Delta U (\text{Case-}I) = \Delta U (\text{Case-}II) = \Delta U (\text{Case-}III) = 0$.

(C) Statement $I$ is true because during the phase transition of ice to water at its melting point, the temperature remains constant at $0 \ ^\circ C$ until all the ice has melted, as the heat supplied is used as latent heat of fusion.
Statement $II$ is also true because the absorbed heat is utilized to overcome the intermolecular forces of attraction (hydrogen bonding) between water molecules in the solid lattice, and since the temperature does not change, the average kinetic energy of the molecules remains constant.
Therefore, both statements are correct.

(B) The process from $1$ to $4$ consists of three steps: $1 \rightarrow 2$,$2 \rightarrow 3$,and $3 \rightarrow 4$.
Step $1 \rightarrow 2$: Isochoric process $(V = 1000 \ cm^3 = 1 \ L)$,so $W_{1 \rightarrow 2} = 0 \ J$.
Step $2 \rightarrow 3$: Isobaric expansion $(P = 3 \ atm)$ from $V_2 = 1 \ L$ to $V_3 = 2 \ L$.
$W_{2 \rightarrow 3} = -P \Delta V = -3 \ atm \times (2 \ L - 1 \ L) = -3 \ L \ atm$.
Step $3 \rightarrow 4$: Isochoric process $(V = 2000 \ cm^3 = 2 \ L)$,so $W_{3 \rightarrow 4} = 0 \ J$.
Total work done $W = W_{1 \to 2} + W_{2 \to 3} + W_{3 \to 4} = 0 + (-3 \text{ L atm}) + 0 = -3 \text{ L atm}$
Converting to Joules: $W = -3 \times 101.3 \ J = -303.9 \ J$.
Rounding to the nearest integer,the magnitude is $304 \ J$.

(D) The standard state of a substance is defined as its pure form at a pressure of $1 \ bar$ and a specified temperature (usually $298 \ K$).
$\Delta_{f} H^{\theta}$ (standard enthalpy of formation) is defined as zero for elements in their most stable state at the specified temperature.
For $O_{2(g)}$,which is the most stable form of oxygen,$\Delta_{f} H^{\theta} = 0$ at any specified temperature.
Therefore,$\Delta_{f} H_{500}^{\theta}$ is zero for $O_{2(g)}$ is the correct statement.

(B) 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.
$E$ (Internal energy),$H$ (Enthalpy),and $G$ (Gibbs free energy) are state functions.
$q$ (Heat) and $w$ (Work) are path functions,as their values depend on the process or path taken.
Therefore,the correct set of state functions is $E, H, G$.

(B) Intensive properties are those that do not depend on the amount of matter present in the system.
$pH$,Temperature,Specific heat capacity,Molar internal energy,$E.M.F.$,Molar mass,Vapour density,and Density are intensive properties.
Extensive properties depend on the amount of matter,such as Volume,Mass,and Resistance.
In option $B$,Specific heat capacity,Molar internal energy,and $E.M.F.$ are all intensive properties.
Therefore,the correct option is $B$.

(D) The standard state of a substance is its most stable physical form at a specified temperature (usually $298 \text{ K}$ or $25^{\circ} C$) and $1 \text{ bar}$ (or $1 \text{ atm}$) pressure.
At $25^{\circ} C$ and $1 \text{ atm}$ pressure,water exists as a liquid $(H_{2}O_{(l)})$,not as a gas $(H_{2}O_{(g)})$.
Therefore,$H_{2}O_{(g)}$ is not in its standard state under these conditions.

(A) 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.
$1$. Surface tension is an intensive property because it depends only on the nature of the substance and temperature,not on the quantity of the liquid.
$2$. Volume,internal energy,and the number of moles are extensive properties because they depend on the amount of matter present in the system.

(C) $1$. An intensive property is a property that does not depend on the amount of matter present in the system (e.g.,$Temperature$,$Pressure$,$Density$).
$2$. $A$ 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 (e.g.,$Internal \ energy$,$Enthalpy$,$Entropy$,$Temperature$).
$3$. $Internal \ energy$,$Volume$,and $Entropy$ are extensive properties because they depend on the amount of matter.
$4$. $Temperature$ is independent of the amount of matter (intensive) and its value depends only on the current state of the system (state function).
$5$. Therefore,$Temperature$ is the correct answer.

(D) 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.
$Volume$,$Enthalpy$,and $Entropy$ are extensive properties because they depend on the quantity of matter.
$Molar volume$ is defined as the volume per mole of a substance $(V_m = V/n)$,which is independent of the total amount of the substance.
Therefore,$Molar volume$ is an intensive property.

(A) An intensive property is independent of the amount of substance present in the system. Examples include viscosity,surface tension,and specific heat.
An extensive property depends on the amount of substance present in the system.
Internal energy $(U)$ is an extensive property because its value depends on the quantity of matter in the system.

(A) The correct answer is $A$.
Intensive properties are properties that do not depend on the amount or size of the substance and remain the same regardless of the quantity of the material present.
Extensive properties depend on the amount of substance.
$1$. Surface tension and viscosity are both intensive properties because they are independent of the amount of substance.
$2$. Mass,internal energy,and heat capacity are extensive properties as they depend on the amount of matter.
$3$. Temperature,boiling point,and specific heat are intensive properties.
Since both surface tension and viscosity are intensive,option $A$ is correct.

(C) An intensive property is a physical property of a system that does not depend on the system size or the amount of material in the system.
$Specific \ heat$ and $pressure$ are independent of the amount of matter present,hence they are intensive properties.
In contrast,$mass$,$heat \ capacity$,and $internal \ energy$ are extensive properties as they depend on the amount of matter.

(B) system is called $closed$ if it can exchange energy but not matter with the surroundings.
In a half-filled closed vessel containing boiling water,both liquid water and water vapor are present.
Since there are two distinct phases (liquid and gas) present in the system,it is $heterogeneous$.
Therefore,the system is a $heterogeneous$ $closed$ $system$.

(A) Intensive properties are those which do not depend on the quantity or size of matter present in the system,such as boiling point.
Extensive properties are those whose value depends on the quantity or size of matter present in the system,such as heat capacity.
Therefore,boiling point is an intensive property and heat capacity is an extensive property.

(B) Extensive properties depend on the amount of matter present in the system (e.g.,$Volume$,$Number \ of \ moles$,$Internal \ energy$,$Heat \ capacity$).
Intensive properties are independent of the amount of matter present in the system (e.g.,$Pressure$,$Temperature$,$Surface \ tension$,$Density$).
In option $B$,$Volume$ is an extensive property and $Pressure$ is an intensive property. Therefore,the pair ($Volume$,$Pressure$) matches the requirement.

(D) 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. $Enthalpy$ $(H)$,$Volume$ $(V)$,and $Pressure$ $(P)$ are state functions. $Work$ $(w)$ and $Heat$ $(q)$ are path functions,as their values depend on the process or path followed. Therefore,$Work$ is not a state function.

(A) An intensive property is independent of the amount of matter present in the system.
An extensive property depends on the amount of matter present in the system.
Heat capacity $(C)$ is defined as the product of mass $(m)$ and specific heat capacity $(c)$,i.e.,$C = m \times c$.
Since it depends on the mass of the substance,heat capacity is an extensive property.
Viscosity,pressure,and surface tension are independent of the amount of matter and are therefore intensive properties.

(C) Extensive properties are those properties of a system that depend on the amount of matter (size and mass) present in the system.
$Mass$,$Volume$,and $Internal \ energy$ are extensive properties because they depend on the quantity of matter.
$Pressure$ is an intensive property because it is independent of the amount of matter present in the system.

(B) Intensive properties are those that are independent of the amount of matter present in the system.
Extensive properties depend on the quantity of matter present.
$1$. Surface tension: Intensive property.
$2$. Heat capacity: Extensive property (as it depends on the mass of the substance).
$3$. Viscosity: Intensive property.
$4$. Temperature: Intensive property.
Therefore,heat capacity is $NOT$ an intensive property.

(D) $I$. Extensive property: Heat capacity (depends on the amount of matter present in the system).
$II$. Intensive property: Surface tension,density,and refractive index (independent of the amount of matter present in the system).

(A) An extensive property is a property whose value depends on the quantity or size of matter present in the system (e.g.,heat,mass,volume).
An intensive property is a property whose value is independent of the quantity or size of matter present in the system (e.g.,temperature,pressure,density).
Therefore,heat is an extensive property and temperature is an intensive property.

(C) State functions are properties that depend only on the initial and final states of the system,such as internal energy $(U)$,enthalpy $(H)$,and entropy $(S)$.
Path functions are properties that depend on the path taken to reach the state,such as work $(w)$ and heat $(q)$.

(A) Work done in a chemical reaction at constant pressure is given by $W = -P \Delta V = -\Delta n_g RT$,where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
For zero work,$\Delta n_g$ must be equal to $0$.
In option $A$: $\Delta n_g = (1 + 1) - (1 + 1) = 0$.
In option $B$: $\Delta n_g = 2 - (3 + 1) = -2$.
In option $C$: $\Delta n_g = 2 - (1 + 2.5) = -1.5$.
In option $D$: $\Delta n_g = 4 - (2 + 7) = -5$.
Since $\Delta n_g = 0$ for reaction $A$,it performs zero work.

(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ext} \Delta V$.
Given: $W = -600 \ J$,$V_1 = 15 \ dm^{3}$,$V_2 = 20 \ dm^{3}$.
Change in volume $\Delta V = V_2 - V_1 = 20 \ dm^{3} - 15 \ dm^{3} = 5 \ dm^{3}$.
Since $1 \ dm^{3} = 1 \ L$ and $1 \ L \cdot bar = 100 \ J$,we convert the volume change to units compatible with Joules.
$5 \ dm^{3} = 5 \ L$.
Using $W = -P_{ext} \Delta V$,we have $-600 \ J = -P_{ext} \times (5 \ L)$.
$P_{ext} = \frac{600 \ J}{5 \ L} = 120 \ J/L$.
Since $100 \ J = 1 \ L \cdot bar$,then $120 \ J = 1.2 \ L \cdot bar$.
Therefore,$P_{ext} = \frac{120 \ J}{5 \ L} = 24 \ J/L = 0.24 \ bar$.
Wait,re-evaluating the conversion: $1 \ dm^{3} \cdot bar = 100 \ J$.
$P_{ext} (bar) = \frac{-W (J)}{100 \times \Delta V (dm^{3})} = \frac{600}{100 \times 5} = \frac{600}{500} = 1.2 \ bar$.
The correct answer is $1.2 \ bar$.

(D) The work done in a thermodynamic process is given by the formula $W = -P_{ext} \Delta V$.
In the case of free expansion,a gas expands against a vacuum,meaning the external pressure $P_{ext} = 0$.
Therefore,the work done $W = -0 \times \Delta V = 0$.
Thus,free expansion involves zero work done.

(A) The work done during the expansion of a gas against a constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 1 \ atm$,$V_1 = 15.5 \ dm^3$,and $V_2 = 20 \ dm^3$.
Change in volume,$\Delta V = V_2 - V_1 = 20 \ dm^3 - 15.5 \ dm^3 = 4.5 \ dm^3$.
Since $1 \ dm^3 \cdot atm = 101.325 \ J$,the work done is $W = -1 \ atm \times 4.5 \ dm^3 = -4.5 \ dm^3 \cdot atm$.
Converting to joules: $W = -4.5 \times 101.325 \ J \approx -456 \ J$.

(C) For an isothermal reversible expansion,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log\left(\frac{P_1}{P_2}\right)$.
Given values are: $n = 1 \ mol$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$P_1 = 10 \ bar$,and $P_2 = 1 \ bar$.
Substituting these values into the equation:
$w = -2.303 \times 1 \times 8.314 \times 300 \times \log\left(\frac{10}{1}\right)$.
Since $\log(10) = 1$,we have:
$w = -2.303 \times 8.314 \times 300 \times 1$.
$w = -5744.14 \ J$.
Converting to $kJ$,we get $w = -5.744 \ kJ \approx -5.74 \ kJ$.

(A) The work done in an irreversible isothermal expansion against constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Given:
$P_{ext} = 1 \ atm$
$V_1 = 2 \ dm^3$
$V_2 = 2.8 \ dm^3$
$\Delta V = V_2 - V_1 = 2.8 \ dm^3 - 2 \ dm^3 = 0.8 \ dm^3$.
Since $1 \ dm^3 \cdot atm = 101.325 \ J$,
$W = -1 \ atm \times 0.8 \ dm^3 = -0.8 \ dm^3 \cdot atm$.
$W = -0.8 \times 101.325 \ J = -81.06 \ J$.
Rounding to the nearest option,the work done is $-81.04 \ J$.

(B) The work done in an irreversible process against constant external pressure is given by the formula: $W = -P_{ext} \times \Delta V$.
Here,$P_{ext} = 4 \ bar = 4 \times 10^5 \ Pa$.
Initial volume $V_1 = 25 \ dm^3 = 25 \times 10^{-3} \ m^3$.
Final volume $V_2 = 13 \ dm^3 = 13 \times 10^{-3} \ m^3$.
Change in volume $\Delta V = V_2 - V_1 = (13 - 25) \times 10^{-3} \ m^3 = -12 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(4 \times 10^5 \ Pa) \times (-12 \times 10^{-3} \ m^3) = 4800 \ J$.
Since the gas is compressed,work is done on the system,so the value is positive.

$(A)$ The formula for work done in an irreversible process against constant external pressure is given by $W = -P_{ext} \times \Delta V$.
Here, $P_{ext} = 3 \ bar = 3 \times 10^5 \ Pa$.
$\Delta V = V_f - V_i = 13 \ dm^3 - 24 \ dm^3 = -11 \ dm^3 = -11 \times 10^{-3} \ m^3$.
Substituting the values: $W = -(3 \times 10^5 \ Pa) \times (-11 \times 10^{-3} \ m^3)$.
$W = 3300 \ J$.
Since the gas is compressed, work is done on the system, hence the value is positive.