Isobaric and Isochoric Processes Questions in English

(A) For an isobaric process,the work done $W$ is given by $W = P \Delta V = nR \Delta T$.
The heat supplied $Q$ at constant pressure is given by $Q = n C_p \Delta T$.
Using Mayer's relation,$C_p = C_v + R$ (where $C_v$ and $C_p$ are molar heat capacities).
Therefore,the heat supplied is $Q = n(C_v + R) \Delta T = (n C_v + nR) \Delta T$.
The ratio of work done to heat supplied is $\frac{W}{Q} = \frac{nR \Delta T}{(n C_v + nR) \Delta T} = \frac{nR}{n C_v + nR}$.
If $C_v$ is the total heat capacity of $n$ moles,then $C_v(\text{total}) = n C_v(\text{molar})$. Thus,the ratio is $\frac{nR}{C_v + nR}$.

(C) For a diatomic gas with rigid molecules,the molar heat capacity at constant pressure is given by $C_{p} = \frac{7}{2} R$.
The work done in an isobaric process is given by $W = P \Delta V = nR \Delta T = 10 \ J$.
The heat energy absorbed by the gas is given by $\Delta Q = n C_{p} \Delta T$.
Substituting $C_{p} = \frac{7}{2} R$,we get $\Delta Q = n \left( \frac{7}{2} R \right) \Delta T = \frac{7}{2} (nR \Delta T)$.
Since $nR \Delta T = W = 10 \ J$,we have $\Delta Q = \frac{7}{2} \times 10 \ J = 35 \ J$.

(C) For an ideal gas at constant pressure,the heat supplied $Q$ is given by $Q = n C_p \Delta T$.
The work done by the gas is $W = P \Delta V = n R \Delta T$.
We know that the molar heat capacity at constant pressure is $C_p = \frac{\gamma R}{\gamma - 1}$.
Substituting this into the heat equation: $Q = n \left( \frac{\gamma R}{\gamma - 1} \right) \Delta T$.
From the work equation,we have $n R \Delta T = W$. Substituting this into the heat equation:
$Q = W \left( \frac{\gamma}{\gamma - 1} \right)$.
Rearranging to solve for $W$:
$W = Q \left( \frac{\gamma - 1}{\gamma} \right)$.
Given $Q = 100 \ J$ and $\gamma = 1.4$:
$W = 100 \left( \frac{1.4 - 1}{1.4} \right) = 100 \left( \frac{0.4}{1.4} \right) = 100 \left( \frac{4}{14} \right) = 100 \left( \frac{2}{7} \right) \approx 28.57 \ J$.

(C) For a monoatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{5}{2} R$ and at constant volume is $C_v = \frac{3}{2} R$.
The heat supplied at constant pressure is given by $dQ = n C_p dT = n \left( \frac{5}{2} R \right) dT$.
The work done by the gas at constant pressure is $dW = P dV = n R dT$.
Dividing the work done by the heat supplied,we get:
$\frac{dW}{dQ} = \frac{n R dT}{n (\frac{5}{2} R) dT} = \frac{1}{5/2} = \frac{2}{5}$.
Therefore,the work done by the gas is $dW = \frac{2}{5} Q$.

(N/A) Consider a gas enclosed in a cylinder fitted with a frictionless,movable piston of cross-sectional area $A$. When the gas expands at constant pressure $P$,the piston moves outward by a small distance $\Delta x$.
The force exerted by the gas on the piston is $F = P \times A$.
The work done by the gas during this small displacement $\Delta x$ is:
$\Delta W = F \times \Delta x = (P \times A) \times \Delta x$
Since the change in volume is $\Delta V = A \times \Delta x$,we can write:
$\Delta W = P \Delta V$
For a finite change in volume from $V_i$ to $V_f$ at constant pressure $P$,the total work done is:
$W = \int_{V_i}^{V_f} P \, dV = P(V_f - V_i) = P \Delta V$
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Substituting the expression for work,we get:
$\Delta Q = \Delta U + P \Delta V$

(A) In an isobaric process,the pressure $P$ of the system remains constant throughout the process.
Work done $W$ by a gas is defined by the integral $W = \int_{V_i}^{V_f} P \, dV$.
Since $P$ is constant,it can be taken out of the integral: $W = P \int_{V_i}^{V_f} dV$.
Evaluating the integral,we get $W = P [V]_{V_i}^{V_f}$.
Thus,the work done is $W = P(V_f - V_i)$,where $V_f$ is the final volume and $V_i$ is the initial volume.

(A) In a $P-V$ diagram:
$1$. $A$ horizontal line parallel to the $V$-axis indicates that the pressure $P$ remains constant as the volume $V$ changes. This represents an isobaric process. Thus,$(a)$ corresponds to $(ii)$.
$2$. $A$ vertical line parallel to the $P$-axis indicates that the volume $V$ remains constant as the pressure $P$ changes. This represents an isochoric process. Thus,$(b)$ corresponds to $(iii)$.

(N/A) The process during which the volume of a system remains constant is known as an isochoric process. In an isochoric process,the volume $V$ is constant.
Since work $W = P \Delta V$ and $\Delta V = 0$,the work done $W = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta W = 0$,we have $\Delta Q = \Delta U$.
This implies that if heat is absorbed,the internal energy increases,leading to an increase in temperature. Conversely,if the gas rejects heat,the internal energy decreases,leading to a decrease in temperature.
The change in temperature for a given amount of heat can be determined by the specific heat capacity of the gas at constant volume using the formula:
$C_v = \frac{\Delta Q}{\mu \Delta T} \implies \Delta T = \frac{\Delta Q}{\mu C_v}$,where $\mu$ is the number of moles.

(N/A) An isobaric process is a thermodynamic process in which the pressure of the system remains constant.
Consider a gas undergoing an expansion from an initial state $(V_1, P)$ to a final state $(V_2, P)$ at a constant pressure $P$,as shown in the $P-V$ diagram.
The work done $W$ by the gas is given by the integral of pressure with respect to volume:
$W = \int_{V_1}^{V_2} P \, dV$
Since the pressure $P$ is constant,it can be taken out of the integral:
$W = P \int_{V_1}^{V_2} dV$
$W = P [V_2 - V_1]$
$W = P \Delta V$
Using the ideal gas equation $PV = \mu RT$,we can express the work done in terms of temperature change:
$PV_1 = \mu RT_1$
$PV_2 = \mu RT_2$
Substituting these into the work equation:
$W = \mu RT_2 - \mu RT_1$
$W = \mu R(T_2 - T_1)$
$W = \mu R \Delta T$

(N/A) An isobaric process is a thermodynamic process in which the pressure $P$ of the system remains constant.
For an ideal gas,the work done $W$ during a process is given by the integral $W = \int_{V_i}^{V_f} P \, dV$.
Since the pressure $P$ is constant in an isobaric process,it can be taken out of the integral:
$W = P \int_{V_i}^{V_f} dV$
$W = P(V_f - V_i)$
Using the ideal gas equation $PV = nRT$,we can also express this as:
$W = nR(T_f - T_i)$
where $n$ is the number of moles,$R$ is the universal gas constant,$V_i$ and $V_f$ are the initial and final volumes,and $T_i$ and $T_f$ are the initial and final temperatures.

(N/A) In an isobaric process,the pressure $P$ remains constant. Therefore,the $P-V$ diagram is a straight line parallel to the volume axis ($V$-axis).
In an isochoric process,the volume $V$ remains constant. Therefore,the $P-V$ diagram is a straight line parallel to the pressure axis ($P$-axis).

(A) In a $P-V$ diagram:
$1$. Graph $(a)$ is a horizontal line,which means pressure $P$ remains constant as volume $V$ changes. This represents an Isobaric process.
$2$. Graph $(b)$ is a vertical line,which means volume $V$ remains constant as pressure $P$ changes. This represents an Isochoric process.
Therefore,the correct matching is: $(a-ii), (b-iii)$.

(D) In a $P-V$ diagram,the $y$-axis represents pressure $(P)$ and the $x$-axis represents volume $(V)$.
From the given figure,the line representing the process is horizontal,which means the pressure $(P)$ remains constant as the volume $(V)$ changes from the initial state to the final state.
$A$ thermodynamic process in which the pressure of the system remains constant is known as an isobaric process.
Therefore,the correct option is $D$.

(A) For an isobaric process,the work done $W$ and heat supplied $Q$ are given by:
$W = n R \Delta T$
$Q = n C_p \Delta T$
Since $C_p = \frac{f}{2} R + R = (\frac{f}{2} + 1) R$,we have:
$\frac{W}{Q} = \frac{n R \Delta T}{n (\frac{f}{2} + 1) R \Delta T} = \frac{1}{\frac{f}{2} + 1}$
For a diatomic gas,the degrees of freedom $f = 5$.
Substituting $f = 5$:
$\frac{W}{Q} = \frac{1}{\frac{5}{2} + 1} = \frac{1}{\frac{7}{2}} = \frac{2}{7}$
Therefore,$Q = \frac{7}{2} W$.
Given $W = 10 \ J$,we get:
$Q = \frac{7}{2} \times 10 = 35 \ J$.

(B) For an ideal gas,the change in internal energy is given by $dU = n C_{v} dT$.
The heat supplied at constant pressure is given by $dQ = n C_{p} dT$.
The work done in an isobaric process is given by $dW = P dV = n R dT$.
Therefore,the ratio $dU : dQ : dW$ is $n C_{v} dT : n C_{p} dT : n R dT$,which simplifies to $C_{v} : C_{p} : R$.
Substituting the given values,we get $\frac{5}{2} R : \frac{7}{2} R : R$.
Multiplying by $\frac{2}{R}$,we obtain the ratio $5 : 7 : 2$.

(A) Since the pressure of the gas remains constant,the process is isobaric.
For a diatomic gas where molecules rotate but do not oscillate,the degrees of freedom $f = 5$.
The change in internal energy is given by $\Delta U = n C_v \Delta T = n \left( \frac{f}{2} R \right) \Delta T = \frac{5}{2} n R \Delta T$.
The work done by the gas in an isobaric process is $W = P \Delta V = n R \Delta T$.
The ratio of the change in internal energy to the work done is $\frac{\Delta U}{W} = \frac{\frac{5}{2} n R \Delta T}{n R \Delta T} = \frac{5}{2} = 2.5$.
Given that this ratio is equal to $\frac{x}{10}$,we have $\frac{x}{10} = 2.5$.
Therefore,$x = 2.5 \times 10 = 25$.

(D) For an isobaric process,the work done is given by $W = P \Delta V = nR \Delta T = 400 \ J$.
The heat supplied to the gas is given by $Q = nC_p \Delta T$.
Since $C_p = \frac{\gamma R}{\gamma - 1}$,we have $Q = n \left( \frac{\gamma R}{\gamma - 1} \right) \Delta T$.
Substituting $nR \Delta T = W = 400 \ J$ and $\gamma = 1.4$:
$Q = \frac{\gamma}{\gamma - 1} \times W = \frac{1.4}{1.4 - 1} \times 400$.
$Q = \frac{1.4}{0.4} \times 400 = 3.5 \times 400 = 1400 \ J$.

(C) The heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$.
The work done by the gas at constant pressure is $\Delta W = p \Delta V = n R \Delta T$.
The ratio of the work done to the heat supplied is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.
Using the relation $R = C_p - C_V$,we get $\frac{\Delta W}{\Delta Q} = \frac{C_p - C_V}{C_p} = 1 - \frac{C_V}{C_p} = 1 - \frac{1}{\gamma}$.
For a diatomic gas,the adiabatic index $\gamma = 7/5$.
Therefore,the ratio is $1 - \frac{1}{7/5} = 1 - \frac{5}{7} = \frac{2}{7}$.

(D) Cylinder $A$ has a free piston,which means the process occurs at constant pressure. The heat supplied is given by $\Delta Q = n C_P \Delta T_A = n C_P T_0$.
Cylinder $B$ has a fixed piston,which means the process occurs at constant volume. The heat supplied is given by $\Delta Q = n C_V \Delta T_B$.
Since the same amount of heat is supplied to both cylinders,we have:
$n C_P T_0 = n C_V \Delta T_B$
$\Delta T_B = \frac{C_P}{C_V} T_0$
For an ideal monatomic gas,the ratio of specific heats is $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$.
Therefore,$\Delta T_B = \frac{5}{3} T_0$.

(B) In an isochoric process,the volume of the system remains constant,so the change in volume $\Delta V = 0$.
Since the work done by the system is given by $\Delta W = P \Delta V$,it follows that $\Delta W = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Substituting $\Delta W = 0$ into the equation,we get $\Delta Q = \Delta U$.

(C) From the given graph, the pressure $P$ is directly proportional to the absolute temperature $T$, i.e., $P \propto T$ or $P = kT$ (where $k$ is a constant).
Using the ideal gas equation $PV = nRT$, we can substitute $P = kT$ to get $(kT)V = nRT$, which simplifies to $V = nR/k = \text{constant}$.
Since the volume $V$ remains constant throughout the process, this is an isochoric process.
The molar heat capacity of an ideal gas for an isochoric process is given by $C_V$.
For a diatomic gas, the degrees of freedom $f = 5$.
Therefore, the molar heat capacity $C_V = \frac{f}{2}R = \frac{5}{2}R = 2.5R$.
Thus, the molar heat capacity is $2.5R$.

(C) For a process at constant volume,the work done $W = 0$. According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,so $\Delta Q = \Delta U$.
The change in internal energy is given by $\Delta U = m \cdot c_v \cdot \Delta T$.
Given:
$m = 0.08 \text{ kg}$
$c_v = 0.17 \text{ kcal/kg}^{\circ} \text{C} = 0.17 \times 1000 \text{ cal/kg}^{\circ} \text{C} = 170 \text{ cal/kg}^{\circ} \text{C}$
$\Delta T = 5^{\circ} \text{C}$
$J = 4.18 \text{ J/cal}$
Substituting the values:
$\Delta U = 0.08 \text{ kg} \times 170 \text{ cal/kg}^{\circ} \text{C} \times 5^{\circ} \text{C} \times 4.18 \text{ J/cal}$
$\Delta U = 0.08 \times 170 \times 5 \times 4.18 \text{ J}$
$\Delta U = 68 \times 4.18 \text{ J}$
$\Delta U = 284.24 \text{ J}$
Thus,the change in internal energy is approximately $284 \text{ J}$.

(D) For a diatomic gas, the degrees of freedom $f = 5$.
In an isobaric process, the work done is given by $W = nR\Delta T = 200 \,J$.
The heat supplied to the gas is given by $Q = nC_p\Delta T$.
Since $C_p = \frac{f+2}{2}R$, we have $Q = \left(\frac{f+2}{2}\right) nR\Delta T$.
Substituting the values $f = 5$ and $nR\Delta T = 200 \,J$:
$Q = \left(\frac{5+2}{2}\right) \times 200 = \frac{7}{2} \times 200 = 700 \,J$.

(A) For an isobaric process, the work done is given by $W = P \Delta V = nR \Delta T = 100 \,J$.
The heat supplied to the gas is given by the first law of thermodynamics: $Q = \Delta U + W$.
For a diatomic gas, the degrees of freedom $f = 5$. The change in internal energy is $\Delta U = \frac{f}{2} nR \Delta T = \frac{5}{2} nR \Delta T$.
Substituting $\Delta U$ and $W$ into the heat equation: $Q = \frac{5}{2} nR \Delta T + nR \Delta T = \left(\frac{5}{2} + 1\right) nR \Delta T = \frac{7}{2} nR \Delta T$.
Since $nR \Delta T = 100 \,J$, we have $Q = \frac{7}{2} \times 100 = 350 \,J$.

(D) In a $P-V$ diagram,the work done by a gas is given by the area under the curve,which is $\int P \ dV$.
Along the path $b \to c$,the volume $V$ remains constant at $400 \ cm^3$.
Since there is no change in volume $(dV = 0)$,the process is an isochoric process.
Therefore,the work done by the gas along the path $b \to c$ is $W = P \Delta V = P \times 0 = 0 \ J$.

(D) Since the balloon is fully expandable and requires no energy for expansion,the pressure inside the balloon remains constant (equal to atmospheric pressure). Thus,the process is isobaric.
For an isobaric process,the heat required is given by $\Delta Q = n C_p \Delta T$.
For a monoatomic gas like helium,the degrees of freedom $f = 3$.
The molar heat capacity at constant pressure is $C_p = \frac{f}{2} R + R = \frac{3}{2} R + R = \frac{5}{2} R$.
Given: $n = 2 \ mol$,$\Delta T = 35^{\circ} C - 30^{\circ} C = 5 \ K$,and $R = 8.31 \ J / mol \cdot K$.
Substituting the values:
$\Delta Q = 2 \times \left( \frac{5}{2} \times 8.31 \right) \times 5$
$\Delta Q = 5 \times 8.31 \times 5 = 25 \times 8.31 = 207.75 \ J$.
Rounding to the nearest integer,we get $\Delta Q \approx 208 \ J$.

(B) For an ideal gas,the equation of state is $PV = nRT$.
Given that the pressure $P$ increases linearly with temperature $T$,we have $P = kT$,where $k$ is a constant.
Substituting this into the ideal gas equation: $(kT)V = nRT$.
This simplifies to $V = \frac{nR}{k} = \text{constant}$.
Since the volume $V$ is constant,the process is isochoric.
$1$. Work done $W = \int P dV = 0$ because $dV = 0$. Thus,statement $A$ is correct.
$2$. According to the first law of thermodynamics,$Q = \Delta U + W$. Since $W = 0$,$Q = \Delta U$. Thus,statement $B$ is incorrect.
$3$. Since the volume is constant,statement $C$ is incorrect.
$4$. As the temperature increases,the internal energy $\Delta U = nC_v \Delta T$ increases. Thus,statement $D$ is correct.
$5$. Since the volume is constant,the process is isochoric. Thus,statement $E$ is correct.
Therefore,statements $A, D,$ and $E$ are correct.

(A) For an isobaric process,the heat added is given by $E_1 = n C_p \Delta T$.
The change in internal energy is given by $E_2 = n C_v \Delta T$.
Therefore,the ratio is $\frac{E_1}{E_2} = \frac{n C_p \Delta T}{n C_v \Delta T} = \frac{C_p}{C_v} = \gamma$.
For a monoatomic gas,the degrees of freedom $f = 3$.
The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.
Given $\frac{E_1}{E_2} = \frac{x}{9}$,we have $\frac{5}{3} = \frac{x}{9}$.
Solving for $x$,we get $x = \frac{5 \times 9}{3} = 15$.

(A) For an isobaric process,the heat supplied is given by $\Delta Q = \mu C_p \Delta T$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U = \mu C_v \Delta T$ is the change in internal energy and $\Delta W$ is the work done.
Thus,$\Delta W = \Delta Q - \Delta U = \mu C_p \Delta T - \mu C_v \Delta T = \mu (C_p - C_v) \Delta T$.
The fraction of heat converted into work is $f = \frac{\Delta W}{\Delta Q} = \frac{\mu (C_p - C_v) \Delta T}{\mu C_p \Delta T} = 1 - \frac{C_v}{C_p} = 1 - \frac{1}{\gamma}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$.
Therefore,$f = 1 - \frac{1}{5/3} = 1 - \frac{3}{5} = \frac{2}{5}$.

(D) For a monoatomic gas,the degrees of freedom $f = 3$.
At constant pressure,the molar heat capacity is $C_P = \frac{5}{2}R$ and the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$.
The total heat supplied is $\Delta Q = n C_P \Delta T$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
The fraction of heat used to change internal energy is $\frac{\Delta U}{\Delta Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P}$.
Substituting the values,$\frac{\Delta U}{\Delta Q} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5} = 0.6$.
Therefore,the percentage of total heat used is $0.6 \times 100 = 60 \%$.

(B) For a process at constant pressure,the first law of thermodynamics is given by $Q = \Delta U + W$.
The heat supplied is $Q = n C_p \Delta T$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
The external work done is $W = Q - \Delta U = n C_p \Delta T - n C_v \Delta T = n (C_p - C_v) \Delta T = n R \Delta T$.
For a monoatomic gas,$C_v = \frac{3}{2} R$ and $C_p = \frac{5}{2} R$.
The fraction of heat used for work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{5}{2} R} = \frac{2}{5}$.
Converting to percentage: $\frac{2}{5} \times 100 \% = 40 \%$.

(D) For an ideal gas undergoing a process at constant pressure,the heat supplied $dQ$ is given by $dQ = n C_p dT$.
The work done by the gas is $dW = P dV = n R dT$.
The fraction of heat energy utilized in doing external work is $\frac{dW}{dQ} = \frac{n R dT}{n C_p dT} = \frac{R}{C_p}$.
We know that $C_p = \frac{\gamma R}{\gamma - 1}$.
Substituting this,we get $\frac{dW}{dQ} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma} = 1 - \frac{1}{\gamma}$.
Given $\gamma = \frac{5}{3}$,we have $\frac{dW}{dQ} = 1 - \frac{1}{5/3} = 1 - \frac{3}{5} = \frac{2}{5}$.
Converting to percentage: $\frac{2}{5} \times 100 \% = 40 \%$.

(C) For an isobaric process,the pressure $P$ remains constant.
Heat supplied to the system is given by $Q = n C_{P} \Delta T$.
Work done by the system is given by $W = P \Delta V$.
Using the ideal gas equation $PV = nRT$,for a constant pressure process,$P \Delta V = nR \Delta T$.
Therefore,$W = nR \Delta T$.
The ratio of heat supplied to work done is $\frac{Q}{W} = \frac{n C_{P} \Delta T}{n R \Delta T} = \frac{C_{P}}{R}$.
We know that $C_{P} - C_{V} = R$,so $R = C_{P} - C_{V}$.
Substituting $R$ in the ratio: $\frac{Q}{W} = \frac{C_{P}}{C_{P} - C_{V}}$.
Dividing the numerator and denominator by $C_{V}$: $\frac{Q}{W} = \frac{C_{P}/C_{V}}{(C_{P}/C_{V}) - 1}$.
Since $\frac{C_{P}}{C_{V}} = \gamma$,we get $\frac{Q}{W} = \frac{\gamma}{\gamma - 1}$.

(D) According to the First Law of Thermodynamics,$dQ = dU + dW$,where $dW = P dV$.
For an isochoric process,the volume remains constant,meaning $dV = 0$.
Consequently,the work done $dW = P dV = 0$.
Substituting this into the First Law equation,we get $dQ = dU + 0$,which simplifies to $dQ = dU$.
Therefore,the condition $dQ = dU$ holds true for an isochoric process.

(C) For an ideal gas heated at constant pressure,the heat supplied $(Q_p)$ is given by $Q_p = n C_p \Delta T = 100 \ J$.
The work done by the gas is $W = n R \Delta T$.
We know that $C_p = \frac{\gamma R}{\gamma - 1}$.
Substituting this into the heat equation: $Q_p = n \left( \frac{\gamma R}{\gamma - 1} \right) \Delta T = 100 \ J$.
Therefore,$n R \Delta T = Q_p \left( \frac{\gamma - 1}{\gamma} \right)$.
Given $\gamma = 5/3$,we have $\frac{\gamma - 1}{\gamma} = \frac{5/3 - 1}{5/3} = \frac{2/3}{5/3} = 2/5$.
Thus,$W = 100 \ J \times (2/5) = 40 \ J$.

(D) Under isochoric (constant volume) conditions,the volume change $dV = 0$.
Since work done $dW = P \cdot dV$,it follows that $dW = 0$.
According to the First Law of Thermodynamics,$dQ = dU + dW$.
Substituting $dW = 0$,we get $dQ = dU$.

(D) For an ideal gas,the internal energy is a function of temperature only. When no work is done,the process is isochoric (constant volume),so $\Delta W = 0$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta W = 0$,the heat supplied is equal to the change in internal energy: $\Delta Q = \Delta U = n C_V \Delta T$.
For a monoatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2} R$.
Given $n = 2$ moles,$\Delta T = 373 \ K - 273 \ K = 100 \ K$.
Substituting the values: $\Delta Q = 2 \times \frac{3}{2} R \times 100 = 300 R$.

(D) The work done by a gas during expansion is given by the area under the $P-V$ curve.
For a given change in volume from $V_1$ to $V_2$,the pressure $P$ remains higher in an isobaric process compared to an isothermal or adiabatic process.
Since $W = \int_{V_1}^{V_2} P \, dV$,the area under the curve is largest for the isobaric process.
Therefore,the work done by the gas is maximum when the expansion is isobaric.

(A) The change in internal energy $\Delta U$ for an ideal gas is given by $\Delta U = n C_v \Delta T$.
Using the relation $C_v = \frac{R}{\gamma-1}$,we get $\Delta U = n \left( \frac{R}{\gamma-1} \right) \Delta T$.
From the ideal gas law $PV = nRT$,at constant pressure $P$,we have $P \Delta V = nR \Delta T$.
Substituting this into the internal energy equation: $\Delta U = \frac{P \Delta V}{\gamma-1}$.
Given that the volume changes from $V$ to $2V$,the change in volume is $\Delta V = 2V - V = V$.
Therefore,$\Delta U = \frac{P(V)}{\gamma-1} = \frac{PV}{\gamma-1}$.

(A) In an isobaric process,the pressure of the system remains constant throughout the thermodynamic process. By definition,'iso' means same and 'baric' refers to pressure.

(D) For an isobaric process,the heat supplied is given by $\Delta Q = nC_p \Delta T$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$,where $\Delta U = nC_v \Delta T$ is the change in internal energy and $W$ is the work done.
Thus,$W = \Delta Q - \Delta U = nC_p \Delta T - nC_v \Delta T = n(C_p - C_v) \Delta T$.
Taking the ratio $\frac{W}{\Delta Q}$,we get:
$\frac{W}{\Delta Q} = \frac{n(C_p - C_v) \Delta T}{nC_p \Delta T} = \frac{C_p - C_v}{C_p} = 1 - \frac{C_v}{C_p}$.
Since $\gamma = \frac{C_p}{C_v}$,we have $\frac{C_v}{C_p} = \frac{1}{\gamma}$.
Therefore,$\frac{W}{\Delta Q} = 1 - \frac{1}{\gamma} = \frac{\gamma - 1}{\gamma}$.

(C) For an isobaric process,the heat supplied is given by $Q = n C_p \Delta T$.
The work done by the system is given by $W = P \Delta V = n R \Delta T$.
The ratio of heat supplied to work done is $\frac{Q}{W} = \frac{n C_p \Delta T}{n R \Delta T} = \frac{C_p}{R}$.
We know that for an ideal gas,the molar specific heat at constant pressure is $C_p = \frac{R \gamma}{\gamma - 1}$.
Substituting this into the ratio,we get $\frac{Q}{W} = \frac{R \gamma / (\gamma - 1)}{R} = \frac{\gamma}{\gamma - 1}$.

(D) In an isobaric process, the pressure remains constant.
Heat exchanged is given by $Q = n C_p \Delta T$.
Work done by the system is given by $W = P \Delta V = n R \Delta T$.
Using Mayer's relation, $R = C_p - C_v$.
Therefore, $W = n(C_p - C_v) \Delta T$.
The ratio of work done to heat exchanged is $\frac{W}{Q} = \frac{n(C_p - C_v) \Delta T}{n C_p \Delta T} = \frac{C_p - C_v}{C_p} = 1 - \frac{C_v}{C_p}$.
Since $\gamma = \frac{C_p}{C_v}$, we have $\frac{C_v}{C_p} = \frac{1}{\gamma}$.
Thus, $\frac{W}{Q} = 1 - \frac{1}{\gamma} = \frac{\gamma - 1}{\gamma}$.

$(B)$ An isochoric process is a thermodynamic process in which the volume of the system remains constant $(V = \text{constant})$.
In a pressure-volume $(P-V)$ diagram, a constant volume process is represented by a vertical line, as the volume does not change while the pressure varies.
Looking at the given options, Graph $(B)$ shows a vertical line parallel to the pressure axis, indicating that the volume $V$ is constant for all values of pressure $P$.
Therefore, Graph $(B)$ correctly represents an isochoric process.

(B) For an ideal rigid diatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{7}{2}R$ and at constant volume is $C_v = \frac{5}{2}R$.
In an isobaric process,the heat supplied is $\Delta Q = n C_p \Delta T$ and the work done is $W = P \Delta V = n R \Delta T$.
The ratio of work done to the heat supplied is given by:
$\frac{W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$
Substituting the value of $C_p$:
$\frac{W}{\Delta Q} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$

(C) An isochoric process is a thermodynamic process in which the volume $(V)$ of the system remains constant.
In a $P-V$ diagram,a constant volume process is represented by a vertical line,because for a single value of $V$,the pressure $(P)$ can vary.
Looking at the given options:
Graph $(A)$ shows $P$ proportional to $V$.
Graph $(B)$ shows constant pressure (isobaric).
Graph $(C)$ shows a vertical line,which indicates that $V$ is constant while $P$ changes.
Graph $(D)$ shows a curve.
Therefore,graph $(C)$ represents an isochoric process.

(C) In an isochoric process,the volume of the gas remains constant.
According to Gay-Lussac's law,for a fixed mass of gas at constant volume,the pressure is directly proportional to its absolute temperature $(P \propto T)$.
Therefore,$\frac{P_1}{P_2} = \frac{T_1}{T_2}$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 27 + 273 = 300 \ K$
$T_2 = 127 + 273 = 400 \ K$
Substituting these values into the ratio:
$\frac{P_1}{P_2} = \frac{300}{400} = \frac{3}{4}$.

(A) An isochoric process is a thermodynamic process that occurs at a constant volume.
Since the volume remains constant throughout the process,the change in volume is zero.
Therefore,the equation for an isochoric process is $\Delta V = 0$.

(B) According to the first law of thermodynamics,$\Delta U = Q - W$.
$1$. In an adiabatic process,$Q = 0$ by definition,so it does not satisfy the condition.
$2$. In an isothermal process,the temperature remains constant,which implies $\Delta U = 0$ for an ideal gas,so it does not satisfy the condition.
$3$. In an isochoric process,the volume remains constant,which implies $W = P \Delta V = 0$,so it does not satisfy the condition.
$4$. In an isobaric process,the pressure remains constant. During expansion,the volume changes $(W \neq 0)$,the temperature changes $(\Delta U \neq 0)$,and heat is exchanged with the surroundings $(Q \neq 0)$. Therefore,all three quantities are non-zero.

(B) The work done by or on a gas is given by the formula $\Delta W = P \Delta V$,where $P$ is the pressure and $\Delta V$ is the change in volume.
For the work done to be zero,the change in volume $\Delta V$ must be zero.
This implies that the volume $V$ remains constant throughout the process.
$A$ thermodynamic process in which the volume remains constant is known as an isochoric process.
Therefore,in an isochoric process,the work done is zero.