Isobaric and Isochoric Processes Questions in English

(D) An isobaric process is a thermodynamic process in which the pressure of the system remains constant throughout the process.
By definition,if the pressure $P$ is constant,then the change in pressure $\Delta P$ must be equal to zero.
Therefore,the equation that specifies an isobaric process is $\Delta P=0$.

(A) An isochoric process is defined as a process where the volume of the system remains constant $(dV = 0)$.
Since work done $dW = P dV$,if $dV = 0$,then $dW = 0$. Thus,no work is done in the process.
According to the first law of thermodynamics,$dQ = dU + dW$. Since $dW = 0$,the heat exchanged $(dQ)$ is entirely used to change the internal energy $(dU)$ of the system.
Therefore,statement $(A)$ is incorrect because energy exchanged is not used to do work.

(C) An isobaric process is defined as a thermodynamic process in which the pressure of the system remains constant $(P = \text{constant})$.
According to the ideal gas law, $PV = nRT$. Since $P$ is constant, $V \propto T$.
Therefore, if the volume changes during the process, the temperature of the system must also change.
Statement $(C)$ claims that the temperature remains constant, which is incorrect for an isobaric process; it is a characteristic of an isothermal process.

(B) In an isochoric process,the volume of the system remains constant throughout the process.
On a $p-V$ diagram,where pressure $p$ is plotted on the $y$-axis and volume $V$ is plotted on the $x$-axis,a constant volume process is represented by a vertical line.
This vertical line indicates that for any change in pressure,the value of $V$ remains the same.
Looking at the provided diagrams (not fully shown but implied by standard physics problems of this type),the diagram representing a vertical line corresponds to the isochoric process.
Assuming the standard set of diagrams where $II$ represents the vertical line,the correct option is $II$.

(C) For an ideal gas at constant pressure,Charles's Law states that $ V \propto T $.
Since the volume is doubled $( V_2 = 2V_1 )$,the temperature must also double.
The initial temperature is $ T_1 = 27^{\circ}C = 27 + 273 = 300 \text{ K} $.
The final temperature is $ T_2 = 2 \times 300 \text{ K} = 600 \text{ K} $.
The change in temperature is $ \Delta T = 600 \text{ K} - 300 \text{ K} = 300 \text{ K} $.
For a diatomic gas,the molar heat capacity at constant pressure is $ C_p = \frac{7}{2}R $.
The heat added is given by $ Q = n C_p \Delta T $.
Substituting the values: $ Q = 1 \times \frac{7}{2} R \times 300 = 1050 R $.

(C) According to the first law of thermodynamics,the heat supplied $Q$ is equal to the change in internal energy $\Delta U$ plus the work done $W$: $Q = \Delta U + W$.
For a process at constant pressure,the heat supplied is $Q = n C_p \Delta T$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
The work done is $W = Q - \Delta U = n C_p \Delta T - n C_V \Delta T = n R \Delta T$.
The fraction of heat energy converted into work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.
For a monoatomic ideal gas,$C_p = \frac{5}{2} R$.
Therefore,the fraction is $\frac{W}{Q} = \frac{R}{\frac{5}{2} R} = \frac{2}{5}$.

(A) For a monoatomic gas,the degrees of freedom $f = 3$.
At constant pressure,the heat absorbed is given by $Q = n C_p \Delta T$.
Since $C_p = \frac{f+2}{2} R = \frac{5}{2} R$,we have $Q = \frac{5}{2} n R \Delta T$.
From the ideal gas equation,$p \Delta V = n R \Delta T$.
Substituting this into the heat equation: $Q = \frac{5}{2} p \Delta V$.
Given $Q = 80 \ J$ and $\Delta V = 16 \times 10^{-5} \ m^3$,we have:
$80 = \frac{5}{2} \times p \times 16 \times 10^{-5}$.
$80 = 40 \times 10^{-5} \times p$.
$p = \frac{80}{40 \times 10^{-5}} = 2 \times 10^5 \ Nm^{-2}$.

(A) Given: Heat supplied at constant pressure, $Q_P = 210 \,J$.
For a diatomic gas, the molar heat capacity at constant volume is $C_V = \frac{5}{2}R$.
The molar heat capacity at constant pressure is $C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R$.
We know that $Q_P = n C_P \Delta T = 210 \,J$.
Substituting $C_P$, we get $n (\frac{7}{2}R) \Delta T = 210$.
Thus, $n R \Delta T = 210 \times \frac{2}{7} = 60 \,J$.
The work done by the gas at constant pressure is given by $W = P \Delta V = n R \Delta T$.
Therefore, $W = 60 \,J$.

(A) Given: Work done by the gas, $W = 200 \,J$.
For an ideal monoatomic gas expanding at constant pressure, the heat absorbed is given by $Q = n C_p \Delta T$.
The work done is given by $W = n R \Delta T$.
Taking the ratio of $Q$ to $W$:
$\frac{Q}{W} = \frac{n C_p \Delta T}{n R \Delta T} = \frac{C_p}{R}$.
For a monoatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2} R = 1.5 R$.
Using the relation $C_p = C_v + R$, we get $C_p = 1.5 R + R = 2.5 R$.
Substituting this into the ratio:
$\frac{Q}{W} = \frac{2.5 R}{R} = 2.5$.
Therefore, $Q = 2.5 \times W = 2.5 \times 200 \,J = 500 \,J$.

(B) The change in internal energy $(\Delta U)$ for a gas at constant volume is given by the formula:
$\Delta U = n C_V \Delta T$
Where:
$n = 3 \text{ moles}$
$\Delta U = 1080 \ J$
$\Delta T = 40^{\circ}C - 20^{\circ}C = 20 \ K$
Rearranging the formula to solve for $C_V$:
$C_V = \frac{\Delta U}{n \Delta T}$
Substituting the given values:
$C_V = \frac{1080}{3 \times 20}$
$C_V = \frac{1080}{60}$
$C_V = 18 \ J \ mol^{-1} \ K^{-1}$
Thus, the molar specific heat at constant volume is $18 \ J \ mol^{-1} \ K^{-1}$.

(C) Work done by a gas in a constant pressure process is given by the formula:
$W = P \Delta V$
Since the gas is ideal, we use the ideal gas equation $PV = nRT$. At constant pressure, $P \Delta V = nR \Delta T$.
Therefore, $W = nR \Delta T$ ... $(i)$
Given:
$n = 1.5 \,mol$
$R = 8.3 \,J \,mol^{-1} \,K^{-1}$
$\Delta T = (130 + 273) - (30 + 273) = 100 \,K$
Substituting these values into equation $(i)$:
$W = 1.5 \times 8.3 \times 100$
$W = 1245 \,J$

(B) In the given graph, $V \propto T$, which implies $\frac{V}{T} = \text{constant}$.
This indicates that the process is an isobaric process (constant pressure process).
For an isobaric process, the heat supplied is given by $\Delta Q = n C_p \Delta T$.
The work done is given by $\Delta W = p \Delta V = n R \Delta T$.
Since $C_p = \frac{5}{2} R$ for a monoatomic gas like helium, we have:
$\Delta W = n R \Delta T = n R \left( \frac{\Delta Q}{n C_p} \right) = \frac{\Delta Q R}{C_p} = \frac{\Delta Q R}{\frac{5}{2} R} = \frac{2}{5} \Delta Q$.
Given $\Delta Q = 5 \text{ cal}$, we get:
$\Delta W = \frac{2}{5} \times 5 \text{ cal} = 2 \text{ cal}$.
Converting to Joules, $\Delta W = 2 \times 4.2 \text{ J} = 8.4 \text{ J}$.

(C) For an ideal gas undergoing a process at constant pressure,the work done $W$ is given by the formula $W = nR\Delta T$.
Here,the number of moles $n = 6$.
The universal gas constant $R = 8.31 \ J \ mol^{-1} \ K^{-1}$.
The change in temperature $\Delta T = 20^{\circ} C = 20 \ K$ (since the change in temperature is the same in Celsius and Kelvin scales).
Substituting these values into the formula:
$W = 6 \times 8.31 \times 20$
$W = 120 \times 8.31$
$W = 997.2 \ J$.
Therefore,the correct option is $C$.

(C) The work done by a gas during expansion is given by the area under the $P-V$ curve,$W = \int_{V_1}^{V_2} P \, dV$.
For a given change in volume from $V_1$ to $V_2$,the pressure $P$ must be as high as possible throughout the process to maximize the integral.
In an isobaric process,the pressure remains constant at its initial maximum value.
In other processes like isothermal or adiabatic,the pressure decreases as the volume increases.
Therefore,the area under the $P-V$ curve is largest for an isobaric expansion,making the work done maximum.

(B) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
Since the work done in a thermodynamic process is given by $W = p \cdot \Delta V$,substituting $\Delta V = 0$ gives $W = 0$.

(A) For an isobaric process,the work done is $dW = P dV$ and the heat supplied is $dQ = n C_p dT$.
Using the relation $C_p = \frac{\gamma R}{\gamma - 1}$ and $dW = n R dT$,we have:
$\frac{dW}{dQ} = \frac{nR dT}{n C_p dT} = \frac{R}{C_p} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma}$.
Given $\gamma = 1.4$ and $dW = 300 \ J$,we substitute these values:
$\frac{300}{dQ} = \frac{1.4 - 1}{1.4} = \frac{0.4}{1.4} = \frac{4}{14} = \frac{2}{7}$.
Therefore,$dQ = 300 \times \frac{7}{2} = 150 \times 7 = 1050 \ J$.

(D) For an isobaric process,the work done is given by $W = P \Delta V = nR \Delta T = 200 \ J$.
The heat supplied to the gas is given by $\Delta Q = n C_P \Delta T$.
Given that $C_P = \frac{7}{2} R$,we substitute this into the equation:
$\Delta Q = n \left( \frac{7}{2} R \right) \Delta T = \frac{7}{2} (nR \Delta T)$.
Substituting the value of $nR \Delta T = 200 \ J$:
$\Delta Q = \frac{7}{2} \times 200 \ J = 7 \times 100 \ J = 700 \ J$.

(A) For an isobaric process, the heat supplied is given by $\Delta Q = n C_P \Delta T = 700 \,J$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
For a diatomic gas, the adiabatic index is $\gamma = \frac{C_P}{C_V} = 1.4 = \frac{7}{5}$.
From the first law of thermodynamics, $\Delta Q = \Delta U + W$, so the work done by the gas is $W = \Delta Q - \Delta U$.
Since $\Delta U = n C_V \Delta T = n \left(\frac{C_P}{\gamma}\right) \Delta T = \frac{\Delta Q}{\gamma}$, we have:
$W = \Delta Q - \frac{\Delta Q}{\gamma} = \Delta Q \left(1 - \frac{1}{\gamma}\right)$.
Substituting the values:
$W = 700 \left(1 - \frac{5}{7}\right) = 700 \left(\frac{2}{7}\right) = 200 \,J$.

(C) For an isobaric process, the pressure remains constant $(P = \text{constant})$.
From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.
For a diatomic gas, the molar heat capacities are $C_p = \frac{7}{2}R$ and $C_v = \frac{5}{2}R$.
The heat supplied is $\Delta Q = n C_p \Delta T$.
The work done is $\Delta W = n R \Delta T$.
The fraction of heat converted into work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{7}{2}R} = \frac{2}{7}$.
Converting this to a percentage: $\frac{2}{7} \times 100 \approx 28.6 \%$.
Therefore, the percentage of heat converted into work is $28.6 \%$.

(C) For an isobaric process,the work done is given by $W = P \Delta V = n R \Delta T = 100 J$.
For a monoatomic gas,the molar heat capacity at constant pressure is $C_P = \frac{5}{2} R$.
The heat supplied to the gas is given by $Q = n C_P \Delta T$.
Substituting the value of $C_P$,we get $Q = n (\frac{5}{2} R) \Delta T = \frac{5}{2} (n R \Delta T)$.
Since $n R \Delta T = 100 J$,we have $Q = \frac{5}{2} \times 100 J = 250 J$.

(C) For an ideal gas,the change in internal energy $\Delta U$ is given by $\Delta U = n C_V \Delta T$.
Given $n = 1$ mole,so $\Delta U = C_V \Delta T$.
We know that $C_V = \frac{R}{\gamma-1}$.
Thus,$\Delta U = \frac{R \Delta T}{\gamma-1}$.
From the ideal gas equation at constant pressure $p$,$p V = R T$,so $p \Delta V = R \Delta T$.
Here,the volume changes from $V$ to $2V$,so $\Delta V = 2V - V = V$.
Therefore,$R \Delta T = p V$.
Substituting this into the expression for $\Delta U$:
$\Delta U = \frac{p V}{\gamma-1}$.

(B) The work done by a gas during a constant pressure (isobaric) process is given by the formula: $W = P \Delta V$.
Here, $P = 400 \text{ kPa} = 400 \times 10^3 \text{ Pa}$.
The change in volume $\Delta V$ is equal to the cross-sectional area $A$ multiplied by the displacement $h$.
Given $A = 0.3 \text{ m}^2$ and $h = 10 \text{ cm} = 0.1 \text{ m}$.
Therefore, $\Delta V = A \times h = 0.3 \text{ m}^2 \times 0.1 \text{ m} = 0.03 \text{ m}^3$.
Now, substitute the values into the work formula:
$W = (400 \times 10^3 \text{ Pa}) \times (0.03 \text{ m}^3)$
$W = 400,000 \times 0.03 = 12,000 \text{ J}$.
Converting to kilojoules: $W = 12 \text{ kJ}$.

(C) The process from $P_1$ to $P_2$ occurs at a constant volume $V = 1 \text{ m}^3$.
For an isochoric process,the heat exchanged is given by $\Delta Q = n C_v \Delta T$.
Using the ideal gas law $PV = nRT$,we have $nR \Delta T = V \Delta P$.
Substituting this into the heat equation: $\Delta Q = \frac{C_v}{R} (nR \Delta T) = \frac{C_v}{R} V (P_2 - P_1)$.
Given $n = 10 \text{ mol}$,$C_v = 21 \text{ J/K} \cdot \text{mol}$,$R = 8.3 \text{ J/mol} \cdot \text{K}$,$V = 1 \text{ m}^3$,$P_1 = 21.7 \text{ Pa}$,and $P_2 = 30 \text{ Pa}$.
$\Delta Q = \frac{21}{8.3} \times 1 \times (30 - 21.7) = \frac{21}{8.3} \times 8.3 = 21 \text{ J}$.
Thus,$\alpha = 21$.

(B) For an isobaric process,the work done is given by $W = P \Delta V = nR \Delta T = 100 \ J$.
For a diatomic gas,the degree of freedom $f = 5$.
The molar heat capacity at constant pressure is $C_p = (\frac{f}{2} + 1)R = (\frac{5}{2} + 1)R = \frac{7}{2}R$.
The heat supplied to the gas is $Q = nC_p \Delta T = n(\frac{7}{2}R) \Delta T$.
Substituting $nR \Delta T = 100 \ J$,we get $Q = \frac{7}{2} \times 100 = 350 \ J$.