Isobaric and Isochoric Processes Questions in English

(A) According to the First Law of Thermodynamics $(FLOT)$,the change in internal energy $dU$ is given by $dU = dQ - dW$.
Given that $dW = 0$ (isochoric process) and $dQ < 0$ (heat is released by the system).
Substituting these values,we get $dU = dQ - 0 = dQ$.
Since $dQ < 0$,it follows that $dU < 0$.
For an ideal gas,internal energy $U$ is a function of temperature $T$ only $(U \propto T)$.
Therefore,a decrease in internal energy $(dU < 0)$ implies a decrease in temperature $(dT < 0)$.

(A) The work done $W$ by a gas during an expansion or compression process 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 (isobaric process),it can be taken out of the integral: $W = P \int_{V_1}^{V_2} dV$.
Evaluating the integral,we get: $W = P [V]_{V_1}^{V_2} = P(V_2 - V_1)$.
Therefore,the work done by the gas is $P(V_2 - V_1)$.

(C) In an isobaric process,the pressure remains constant. According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
When heat $\Delta Q$ is supplied to the gas,it expands against the external pressure,so work $\Delta W = P \Delta V$ is done by the gas.
Simultaneously,the temperature of the gas increases,which leads to an increase in the internal energy $\Delta U$ of the gas.
Therefore,both work is done by the gas and the internal energy of the gas increases.

(A) The work done during an isobaric expansion is given by the formula $W = P \Delta V$.
Given:
Pressure $P = 2 \, atm = 2 \times 1.013 \times 10^5 \, Pa \approx 2 \times 10^5 \, Pa$.
Initial volume $V_1 = 50 \, L = 50 \times 10^{-3} \, m^3$.
Final volume $V_2 = 150 \, L = 150 \times 10^{-3} \, m^3$.
Change in volume $\Delta V = V_2 - V_1 = (150 - 50) \times 10^{-3} \, m^3 = 100 \times 10^{-3} \, m^3 = 0.1 \, m^3$.
Substituting the values into the work formula:
$W = (2 \times 10^5 \, Pa) \times (0.1 \, m^3) = 2 \times 10^4 \, J$.
Thus,the correct option is $A$.

(C) For an isobaric process,the work done is given by $W = P \Delta V$.
Using the ideal gas equation $PV = nRT$,we can write $W = nR \Delta T$.
Given:
Number of moles $n = 0.1 \, mol$.
Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \, K$.
Since the pressure is constant,by Charles's Law,$V \propto T$. If the volume doubles $(V_2 = 2V_1)$,the temperature must also double $(T_2 = 2T_1 = 600 \, K)$.
Change in temperature $\Delta T = T_2 - T_1 = 600 - 300 = 300 \, K$.
Substituting the values into the work formula:
$W = 0.1 \times 2 \times 300 = 60 \, cal$.

(B) The work done $W$ by a gas during an isobaric expansion is given by the formula: $W = P \Delta V$.
Given:
Pressure $P = 10^3 \, N/m^2$
Change in volume $\Delta V = 0.25 \, m^3$
Substituting the values into the formula:
$W = 10^3 \, N/m^2 \times 0.25 \, m^3$
$W = 1000 \times 0.25 \, J$
$W = 250 \, J$
Therefore,the work done is $250 \, J$.

(D) The work done by a system at constant pressure (isobaric process) is given by the formula $W = P \Delta V$.
Here,the pressure $P$ is atmospheric pressure,which is $1.013 \times 10^5 \ Pa$.
The change in volume $\Delta V = V_f - V_i = 3.34 \ m^3 - 2 \times 10^{-3} \ m^3 = 3.34 - 0.002 = 3.338 \ m^3$.
Substituting these values into the work formula:
$W = (1.013 \times 10^5 \ Pa) \times (3.338 \ m^3)$
$W \approx 338,137 \ J$
Converting to kilojoules $(kJ)$:
$W \approx 338 \ kJ \approx 340 \ kJ$.
Therefore,the work done by the system is approximately $340 \ kJ$.

(B) In a thermodynamic process,the work done by a gas is equal to the area under the $PV$ curve with respect to the volume axis.
For a given expansion from volume ${V_1}$ to ${V_2}$,the pressure $P$ remains constant in an isobaric process,whereas it decreases in both isothermal and adiabatic processes.
Since the isobaric process maintains the highest pressure throughout the expansion,the area under the $PV$ curve is the largest.
Therefore,the work done follows the order: ${W_{adiabatic}} < {W_{isothermal}} < {W_{isobaric}}$.
Thus,the work done is greatest for an isobaric expansion.

(D) The work done in a thermodynamic process at constant pressure is given by the formula $W = P \Delta V$.
Here,the change in volume is $\Delta V = 2.4 \times 10^{-4} \ m^3$.
The constant pressure is $P = 1 \times 10^5 \ N/m^2$.
Substituting these values into the formula:
$W = (1 \times 10^5 \ N/m^2) \times (2.4 \times 10^{-4} \ m^3)$
$W = 2.4 \times 10^{5-4} \ J$
$W = 2.4 \times 10^1 \ J$
$W = 24 \ J$.
Therefore,the work done is $24 \ J$.

(B) For an isobaric process (constant pressure),the work done is given by $W = P \Delta V$.
Using the ideal gas equation $PV = nRT$,for a constant pressure process,$P \Delta V = nR \Delta T$.
Given: $n = 1 \, mol$,$T_1 = 27^{\circ}C = 300 \, K$,$T_2 = 127^{\circ}C = 400 \, K$.
Change in temperature $\Delta T = T_2 - T_1 = 400 - 300 = 100 \, K$.
Using $R = 1.987 \, cal/mol-K$ (since $C_P$ is given in $cal$),$W = nR \Delta T = 1 \times 1.987 \times 100 = 198.7 \, cal$.
To convert work from calories to Joules,we use $1 \, cal \approx 4.184 \, J$.
$W = 198.7 \times 4.184 \approx 831.4 \, J$.
Comparing with the given options,the closest value is $814 \, J$.

(A) An isochoric process is defined as a thermodynamic process in which the volume of the system remains constant,meaning $\Delta V = 0$.
Since the work done in a reversible process is given by $W = \int P \, dV$,if $\Delta V = 0$,then the work done $\Delta W = 0$.
Therefore,the correct option is $(a)$.

(C) The work done by a gas during a thermodynamic process is given by the formula $W = \int P \, dV$.
In an isochoric process,the volume of the system remains constant,which means the change in volume $\Delta V = 0$.
Since the work done $W = P \Delta V$,substituting $\Delta V = 0$ gives $W = P \times 0 = 0$.
Therefore,the work done in an isochoric process is zero.

(D) An isochoric process is a thermodynamic process in which the volume of the system remains constant $(dV = 0)$.
Since the work done by the system is given by $W = \int P dV$,if the volume is constant,the work done is $zero$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since $W = 0$,all heat energy transferred to the system is absorbed as internal energy $(\Delta Q = \Delta U)$.

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

(C) For an isobaric process,the heat supplied is given by $\Delta Q_P = n C_P \Delta T = Q$.
The change in internal energy is $\Delta U = n C_V \Delta T$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,so $\Delta W = \Delta Q_P - \Delta U$.
Substituting the expressions,$\Delta W = n C_P \Delta T - n C_V \Delta T = n \Delta T (C_P - C_V) = n R \Delta T$.
Since $\Delta Q_P = n C_P \Delta T = Q$,we have $n \Delta T = \frac{Q}{C_P}$.
Thus,$\Delta W = \frac{R}{C_P} Q$.
For a monoatomic gas,$C_V = \frac{3}{2}R$ and $C_P = \frac{5}{2}R$.
Therefore,$\Delta W = \frac{R}{5/2 R} Q = \frac{2}{5} Q$.

(A) For an ideal gas undergoing an isobaric process (constant pressure),the first law of thermodynamics is given by $\Delta Q = \Delta U + \Delta W$.
Here,$\Delta Q = n C_P \Delta T$ is the heat supplied,$\Delta U = n C_V \Delta T$ is the change in internal energy,and $\Delta W$ is the external work done.
The fraction of heat utilized for external work is $\frac{\Delta W}{\Delta Q} = \frac{\Delta Q - \Delta U}{\Delta Q} = 1 - \frac{\Delta U}{\Delta Q}$.
Substituting the expressions,we get $\frac{\Delta W}{\Delta Q} = 1 - \frac{n C_V \Delta T}{n C_P \Delta T} = 1 - \frac{C_V}{C_P} = 1 - \frac{1}{\gamma}$.
Given $\gamma = 5/3$,the fraction is $1 - \frac{1}{5/3} = 1 - 3/5 = 2/5$.
Converting to percentage: $(2/5) \times 100\% = 40\%$.

(D) For an isobaric process,the heat supplied is given by the formula: $dQ = \mu C_P dT$.
Here,$\mu = 1 \, mol$,$dQ = 1163.4 \, J$,and $R = 8.31 \, J \, mol^{-1} \, K^{-1}$.
For a diatomic gas like nitrogen,the molar heat capacity at constant pressure is $C_P = \frac{7}{2} R$.
Substituting the values into the equation: $dT = \frac{dQ}{\mu C_P}$.
$dT = \frac{1163.4}{1 \times (\frac{7}{2} \times 8.31)}$.
$dT = \frac{1163.4 \times 2}{7 \times 8.31} = \frac{2326.8}{58.17} \approx 40 \, K$.
Therefore,the increase in temperature is $40 \, K$.

(B) For process $(1)$,the volume $V$ is constant. Since the change in volume $dV = 0$,the work done $W = \int P dV = 0$.
For process $(2)$,the pressure $P$ is constant. The work done is given by $W = \int_{V_1}^{2V_1} P dV = P [V]_{V_1}^{2V_1} = P(2V_1 - V_1) = PV_1$.
Therefore,the work done in the two processes is $0$ and $PV_1$ respectively.

(C) For an isobaric process (constant pressure), the work done $W$ is given by the formula $W = P \Delta V$.
Using the ideal gas equation $PV = nRT$, for a constant pressure process, $P \Delta V = nR \Delta T$.
Given:
Number of moles $n = 1 \, \text{mol}$.
Change in temperature $\Delta T = 100^{\circ}C - 0^{\circ}C = 100 \, K$.
Gas constant $R = 8.3 \, J/mol \cdot K$.
Substituting these values into the formula:
$W = nR \Delta T = 1 \times 8.3 \times 100$.
$W = 830 \, J = 8.3 \times 10^{2} \, J$.

(C) At constant pressure,according to Charles's Law,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given $V_1 = V$,$V_2 = 2V$,and $T_1 = 27 + 273 = 300 \ K$.
Substituting the values: $\frac{V}{300} = \frac{2V}{T_2}$,which gives $T_2 = 600 \ K$.
The change in temperature is $\Delta T = T_2 - T_1 = 600 - 300 = 300 \ K$.
The work done in an isobaric process is given by $W = P \Delta V = \mu R \Delta T$.
Using $\mu = 0.1 \ mol$ and $R \approx 2 \ cal/(mol \cdot K)$:
$W = (0.1) \times (2) \times (300) = 60 \ cal$.

(C) For an isochoric process,the volume $V$ remains constant.
According to Gay-Lussac's Law,$P \propto T$ or $P_1 / T_1 = P_2 / T_2$.
First,convert the temperatures from Celsius to Kelvin:
$T_1 = 27 + 273 = 300 \, K$
$T_2 = 127 + 273 = 400 \, K$
Now,calculate the ratio:
$P_1 / P_2 = T_1 / T_2 = 300 / 400 = 3/4$.

(A) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
Since the work done 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 $\Delta W = 0$,we get $\Delta Q = \Delta U$.

(A) For an isochoric process,the volume remains constant,so $\Delta V = 0$.
From the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
Since $\Delta W = P \Delta V$,and $\Delta V = 0$,we have $\Delta W = 0$.
Therefore,$\Delta Q = \Delta U + 0$,which gives $\Delta Q = \Delta U$.

(B) The work done by a gas during expansion at constant pressure is given by the formula: $\Delta W = P \Delta V$.
Given:
Pressure $P = 10^{3} \ N/m^{2}$
Change in volume $\Delta V = 0.25 \ m^{3}$
Substituting the values:
$\Delta W = 10^{3} \times 0.25 = 250 \ J$.
Therefore,the work done is $250 \ J$.

(A) Given that $C_P/C_V = \gamma$.
We know that $C_P - C_V = R$,so $C_V = \frac{R}{\gamma - 1}$.
The change in internal energy for any process is given by $\Delta U = n C_V \Delta T$.
Since $n = 1$,$\Delta U = C_V \Delta T = \frac{R \Delta T}{\gamma - 1}$.
From the ideal gas equation $PV = nRT$,at constant pressure $P$,$P \Delta V = nR \Delta T$. Since $n = 1$,$P \Delta V = R \Delta T$.
Substituting this into the internal energy equation: $\Delta U = \frac{P \Delta V}{\gamma - 1}$.
Given $\Delta V = 2V - V = V$,we get $\Delta U = \frac{PV}{\gamma - 1}$.

(D) For container $A$ (isobaric process),the heat supplied is given by $\Delta Q_A = \mu C_p \Delta T_A$.
For container $B$ (isochoric process),the heat supplied is given by $\Delta Q_B = \mu C_v \Delta T_B$.
Since the heat supplied is the same,$\Delta Q_A = \Delta Q_B$.
Therefore,$\mu C_p \Delta T_A = \mu C_v \Delta T_B$.
This simplifies to $\Delta T_B = \frac{C_p}{C_v} \Delta T_A = \gamma \Delta T_A$.
For a diatomic gas,the adiabatic index $\gamma = \frac{7}{5} = 1.4$.
Given $\Delta T_A = 30\, K$,we have $\Delta T_B = 1.4 \times 30 = 42\, K$.

(B) For a gas expanding at constant pressure,the heat supplied is $dQ = n C_p dT$.
The work done is $dW = P dV = n R dT$.
The fraction of heat converted into work is $\eta = \frac{dW}{dQ} = \frac{n R dT}{n C_p dT} = \frac{R}{C_p}$.
Since $C_p = \frac{\gamma R}{\gamma - 1}$,we have $\eta = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma} = 1 - \frac{1}{\gamma}$.

(A) For an isobaric process, the heat supplied is $dQ = nC_p dT$.
The change in internal energy is $dU = nC_v dT$.
The work done is $dW = PdV = nR dT$.
The fraction of heat converted to internal energy is $\frac{dU}{dQ} = \frac{nC_v dT}{nC_p dT} = \frac{C_v}{C_p} = \frac{1}{\gamma}$.
Given $\gamma = 5/3$, the fraction is $\frac{1}{5/3} = 3/5 = 0.6$ or $60\%$.
The fraction of heat converted to work is $\frac{dW}{dQ} = \frac{nR dT}{nC_p dT} = \frac{R}{C_p} = \frac{C_p - C_v}{C_p} = 1 - \frac{1}{\gamma} = 1 - 0.6 = 0.4$ or $40\%$.
Thus, the percentages are $60\%$ and $40\%$ respectively.

(C) Since no work is done in the process,the process is isochoric (constant volume).
For an isochoric process,the heat supplied is equal to the change in internal energy: $\Delta Q = \Delta U$.
The formula for internal energy change is $\Delta U = n \, C_v \, \Delta T$.
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: $\Delta Q = 2 \times \frac{3}{2} R \times 100 = 300 \ R$.

(B) The work done $W$ by a gas at constant pressure $P$ is given by the formula $W = P \Delta V$.
Given:
Pressure $P = 10^3\,N/m^2$
Change in volume $\Delta V = 0.25\,m^3$
Substituting the values into the formula:
$W = 10^3\,N/m^2 \times 0.25\,m^3$
$W = 1000 \times 0.25\,J$
$W = 250\,J$
Therefore,the work done is $250\,J$.

(A) From the graph,$V \propto T$,which implies the process is isobaric (constant pressure).
For an isobaric process,the heat absorbed is given by $dQ = nC_p dT$,where $C_p$ is the molar heat capacity at constant pressure.
For a monatomic gas,the degrees of freedom $f = 3$. Thus,$C_p = \frac{f+2}{2}R = \frac{3+2}{2}R = \frac{5}{2}R$.
Therefore,$dQ = n \left( \frac{5}{2}R \right) dT$.
The work done by the gas is $dW = PdV$. Since $PV = nRT$,for a constant pressure process,$PdV = nRdT$.
The required ratio is $\frac{dW}{dQ} = \frac{nRdT}{n(\frac{5}{2}R)dT} = \frac{1}{5/2} = \frac{2}{5}$.

(D) As we know,$1 \ mol$ of any ideal gas at $STP$ occupies a volume of $22.4 \ L$.
Hence,the number of moles of gas $\mu = \frac{44.8}{22.4} = 2 \ mol$.
Since the volume of the cylinder is fixed,the process is isochoric (constant volume).
The heat required at constant volume is given by $\Delta Q_V = \mu C_V \Delta T$.
For a monatomic gas,the molar heat capacity at constant volume is $C_V = \frac{3}{2}R$.
Substituting the values: $\Delta Q_V = 2 \times \frac{3}{2}R \times 10 = 30R$.

(B) The $V-T$ graph is a straight line passing through the origin. According to the ideal gas law $PV = nRT$,if $V \propto T$,then the pressure $P$ remains constant. Thus,the process $1-2$ is an isobaric process.
For an isobaric process,the heat absorbed is given by $\Delta Q = nC_P \Delta T$.
The work done by the gas is given by $\Delta W = P \Delta V = nR \Delta T$.
Using the relation $C_P - C_V = R$,we can write $\Delta W = n(C_P - C_V) \Delta T$.
The ratio of heat absorbed to the work done is:
$\frac{\Delta Q}{\Delta W} = \frac{nC_P \Delta T}{n(C_P - C_V) \Delta T} = \frac{C_P}{C_P - C_V} = \frac{1}{1 - \frac{C_V}{C_P}} = \frac{1}{1 - \frac{1}{\gamma}}$
For a monoatomic gas like helium,the adiabatic index $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$.
Therefore,the ratio is:
$\frac{\Delta Q}{\Delta W} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5$.

(D) From the graph, the process $1 \rightarrow 2$ is an isobaric process because $V \propto T$ (i.e., $V/T = \text{constant}$), which implies pressure $P$ is constant.
For an isobaric process, the heat supplied is given by $\Delta Q = n C_P \Delta T$.
The work done by the gas is $\Delta W = P \Delta V = n R \Delta T$.
We know that for an ideal gas, $C_P = \frac{\gamma R}{\gamma - 1}$.
Substituting this into the expression for $\Delta Q$, we get $\Delta Q = n \left( \frac{\gamma R}{\gamma - 1} \right) \Delta T$.
Now, the ratio $\Delta Q : \Delta W$ is:
$\frac{\Delta Q}{\Delta W} = \frac{n \left( \frac{\gamma R}{\gamma - 1} \right) \Delta T}{n R \Delta T} = \frac{\gamma}{\gamma - 1}$.

(C) The work done by a gas at constant pressure is given by the formula:
$W = P \Delta V$
Given that the pressure $P$ is constant and the volume doubles from an initial volume $V$ to $2V$,the change in volume is $\Delta V = 2V - V = V$.
Substituting this into the work formula:
$W = P(V) = PV$
According to the ideal gas equation,$PV = nRT$. For $n = 1$ mole of gas at temperature $T$,we have:
$PV = RT$
Therefore,the work done by the gas is:
$W = RT$

(D) For an isobaric expansion of an ideal gas,the pressure $P$ is constant.
Since the gas expands,the volume increases,so $\Delta V > 0$.
Work done $W = P \Delta V$,which is not zero.
From the ideal gas equation $PV = nRT$,since $P$ is constant,$V \propto T$. As volume increases,temperature $T$ also increases,so $\Delta U \neq 0$.
From the first law of thermodynamics,$\Delta Q = \Delta U + W$,so $\Delta Q \neq 0$.
For an isobaric process,$V = (nR/P)T$.
Differentiating with respect to $T$,we get $dV/dT = nR/P$,which is a constant.
Differentiating again with respect to $T$,we get $d^2V/dT^2 = 0$.

(C) For an ideal gas at constant pressure,the work done is given by $W = P \Delta V$.
From the ideal gas equation,$PV = nRT$,so for constant pressure,$P \Delta V = nR \Delta T$.
Heat supplied at constant pressure is $Q = n C_p \Delta T$.
For a monoatomic gas,the molar heat capacity at constant pressure is $C_p = \frac{5}{2} R$.
Substituting this into the heat equation,we get $Q = n \left( \frac{5}{2} R \right) \Delta T$.
Since $nR \Delta T = W$,we can write $Q = \frac{5}{2} W$.
Rearranging for work done,we get $W = \frac{2}{5} Q$.

(B) The process from $A$ to $B$ is an isobaric process because the pressure $P$ remains constant at $2 \times 10^5 \text{ Pa}$.
For an ideal gas,the work done by the gas is given by $W_{\text{by}} = nR\Delta T$.
Here,$n = 2 \text{ moles}$,$T_A = 300 \text{ K}$,and $T_B = 500 \text{ K}$.
$W_{\text{by}} = 2 \times R \times (500 - 300) = 400R$.
The question asks for the work done $ON$ the gas.
Work done on the gas is $W_{\text{on}} = -W_{\text{by}} = -400R$.
However,in the context of such physics problems,the magnitude is often expected. Given the options,the magnitude $400R$ is the correct answer.

(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$,which implies $V \propto R^3$.
When the diameter doubles,the radius $R$ also doubles $(R \rightarrow 2R)$.
Therefore,the new volume $V'$ becomes $V' = \frac{4}{3} \pi (2R)^3 = 8 \times (\frac{4}{3} \pi R^3) = 8V$.
The work done against the constant atmospheric pressure $p$ during an isobaric expansion is given by $W = p \Delta V$.
Substituting the values,$W = p(V' - V) = p(8V - V) = 7pV$.

(A) For an isobaric process,the work done $W$ is given by $W = nR\Delta T$ and the heat supplied $Q$ is given by $Q = nC_p\Delta T$.
From the graph,at $Q = 80 \ J$,$W = 32 \ J$.
Therefore,the ratio $\frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p} = \frac{32}{80} = \frac{2}{5}$.
This implies $5R = 2C_p$.
Since $C_p = C_v + R$,we have $5R = 2(C_v + R) = 2C_v + 2R$.
$3R = 2C_v \implies C_v = \frac{3}{2}R$.
$A$ gas with $C_v = \frac{3}{2}R$ is a monoatomic gas. Among the given options,$He$ (Helium) is a monoatomic gas. Thus,the correct option is $A$.

(D) For an ideal gas,the heat supplied at constant pressure is given by $\Delta Q = n C_p \Delta T$.
The change in internal energy is given by $\Delta U = n C_v \Delta T$.
According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,so the work done is $\Delta W = \Delta Q - \Delta U$.
Substituting the expressions,$\Delta W = n C_p \Delta T - n C_v \Delta T = n(C_p - C_v) \Delta T = n R \Delta T$.
The fraction of heat used as work is $\frac{\Delta W}{\Delta Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.
For a monoatomic gas,$C_p = \frac{5}{2} R$.
Therefore,the fraction is $\frac{R}{\frac{5}{2} R} = \frac{2}{5} = 0.4$.

(B) For an ideal gas at constant pressure,the work done $W$ is given by the formula $W = P \Delta V$.
Using the ideal gas equation $PV = nRT$,at constant pressure $P$,we have $P \Delta V = nR \Delta T$.
Given:
Number of moles $n = 1 \ mol$.
Initial temperature $T_1 = 27 \ ^oC = 27 + 273 = 300 \ K$.
Final temperature $T_2 = 127 \ ^oC = 127 + 273 = 400 \ K$.
Change in temperature $\Delta T = T_2 - T_1 = 400 \ K - 300 \ K = 100 \ K$.
Universal gas constant $R = 8.31 \ J/mol-K$.
Substituting the values into the work formula:
$W = nR \Delta T = 1 \times 8.31 \times 100 = 831 \ J$.
Thus,the work done is $831 \ J$.

(C) From the ideal gas law,we know that $PV = nRT$,which implies $\frac{P}{T} = \frac{nR}{V}$.
Substituting this into the given equation $V = K \left( \frac{P}{T} \right)^{0.33}$,we get:
$V = K \left( \frac{nR}{V} \right)^{0.33}$
$V = K (nR)^{0.33} \cdot V^{-0.33}$
Multiplying both sides by $V^{0.33}$,we get:
$V^{1.33} = K' \text{ (where } K' = K(nR)^{0.33} \text{ is a constant)}$
Since $V^{1.33}$ is constant,it implies that $V$ must be a constant.
$A$ process in which the volume remains constant is known as an isochoric process.

(A) For process $AB$,the volume is constant $(V = V_0)$,so it is an isochoric process.
Work done $W_{AB} = 0$.
The heat given is $\Delta Q = \Delta U = n C_v \Delta T$.
For a diatomic gas,$C_v = \frac{5}{2} R$.
Using the ideal gas law $PV = nRT$,we have $n \Delta T = \frac{\Delta(PV)}{R} = \frac{P_B V_B - P_A V_A}{R}$.
Given $P_A = P_0, V_A = V_0$ and $P_B = 2P_0, V_B = V_0$.
So,$n \Delta T = \frac{(2P_0)(V_0) - (P_0)(V_0)}{R} = \frac{P_0 V_0}{R}$.
Substituting this into the heat equation:
$\Delta Q = n \left( \frac{5}{2} R \right) \Delta T = \frac{5}{2} R (n \Delta T) = \frac{5}{2} R \left( \frac{P_0 V_0}{R} \right) = 2.5 P_0 V_0$.

(A) According to the First Law of Thermodynamics,$Q = \Delta U + W$.
For an isobaric process (constant pressure),the heat supplied is $Q = n C_p \Delta T$.
The work done is $W = P \Delta V = n R \Delta T$.
The change in internal energy is $\Delta U = n C_v \Delta T$.
The fraction of heat converted into work is $\frac{W}{Q} = \frac{n R \Delta T}{n C_p \Delta T} = \frac{R}{C_p}$.
Since $C_p = \frac{\gamma R}{\gamma - 1}$,we substitute this into the expression:
$\frac{W}{Q} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma}$.

(C) In a $P-V$ diagram,the work done is equal to the area under the curve.
An isobaric process is one where the pressure remains constant.
In the given figure,the process $AB$ represents an isobaric expansion because the pressure is constant at $P = 2 \times 10^2 \, N/m^2$ while the volume increases from $V_1 = 1 \, m^3$ to $V_2 = 3 \, m^3$.
The work done during an isobaric process is given by $W = P \Delta V$.
Here,$P = 2 \times 10^2 \, N/m^2$ and $\Delta V = V_2 - V_1 = 3 - 1 = 2 \, m^3$.
Therefore,$W = (2 \times 10^2) \times 2 = 400 \, J$.

(A) In an isobaric process,the work done $W$ by an ideal gas is given by $W = P \Delta V = nR \Delta T$.
The heat supplied $Q$ at constant pressure is given by $Q = n C_P \Delta T$.
For a monoatomic gas,the molar heat capacity at constant pressure is $C_P = \frac{5}{2}R$.
The ratio of work done to the heat supplied is $\frac{W}{Q} = \frac{nR \Delta T}{n C_P \Delta T} = \frac{R}{C_P}$.
Substituting the value of $C_P$,we get $\frac{W}{Q} = \frac{R}{\frac{5}{2}R} = \frac{2}{5} = 0.4$.

(A) Given the equation of state: $PV = nRT + \alpha V$.
For $n = 1$ mole,$PV = RT + \alpha V$,which can be rewritten as $V(P - \alpha) = RT$,or $V = \frac{RT}{P - \alpha}$.
Initially,at $T = T_0$ and $P = P_0$,the volume is $V_0 = \frac{RT_0}{P_0 - \alpha}$.
Since the process is isobaric,$P$ remains constant at $P_0$. When the temperature doubles,$T_f = 2T_0$.
The final volume is $V_f = \frac{R(2T_0)}{P_0 - \alpha} = 2V_0$.
The work done in an isobaric process is $W = P_0(V_f - V_0)$.
Substituting the values: $W = P_0(2V_0 - V_0) = P_0 V_0$.
Since $V_0 = \frac{RT_0}{P_0 - \alpha}$,we have $W = P_0 \left( \frac{RT_0}{P_0 - \alpha} \right) = \frac{P_0 T_0 R}{P_0 - \alpha}$.

(B) For an ideal gas undergoing a process at constant pressure (isobaric process),the work done $W$ is given by the formula:
$W = P\Delta V$
Using the ideal gas equation $PV = nRT$,for a constant pressure process,we have $P\Delta V = nR\Delta T$.
Given:
Number of moles $n = 0.5\, mol$
Gas constant $R = 8.31\, J/mol\cdot K$
Change in temperature $\Delta T = 90\,^oC - 20\,^oC = 70\, K$
Substituting the values into the formula:
$W = nR\Delta T$
$W = 0.5 \times 8.31 \times 70$
$W = 290.85\, J$
Rounding to the nearest integer,the work done is approximately $291\, J$.