Thermodynamics

From the graph: A $(2\times10^{-3}\ m^3,\ 2\times10^4$ Pa$)$, B $(2\times10^{-3}\ m^3,\ 6\times10^4$ Pa$)$, C $(4\times10^{-3}\ m^3,\ 6\times10^4$ Pa$)$.
For the cyclic process ABCA, $\Delta U = 0$, so net heat = net work = area enclosed.
$Q_{AB} + Q_{BC} + Q_{CA} = \frac{1}{2}\times(4\times10^{-3} - 2\times10^{-3})\times(6\times10^4 - 2\times10^4)$
$400 + 100 + Q_{CA} = \frac{1}{2}\times(2\times10^{-3})\times(4\times10^4) = 40$ J
$Q_{CA} = -460$ J, so $Q_{AC} = 460$ J

Coefficient of performance $\beta = \frac{Q_2}{W}$, and $\beta = \frac{1-\eta}{\eta}$
$\beta = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = 9$
$Q_2 = \beta W = 9\times 10 = 90$ J

From the graph: A (4 $m^3$, 5 kPa) and B (6 $m^3$, 2 kPa).
$\Delta U = nC_v\Delta T$ with $C_v = \frac{R}{\gamma - 1}$ and $n\Delta T = \frac{P_2V_2 - P_1V_1}{R}$
$\Delta U = \frac{P_2V_2 - P_1V_1}{\gamma - 1}$
For a diatomic gas, $\gamma - 1 = \frac{2}{5}$:
$\Delta U = \frac{(2\times10^3\times 6) - (5\times10^3\times 4)}{2/5} = \frac{-8\times10^3}{2/5}$
$\Delta U = -20$ kJ

$COP = \frac{T_L}{T_H - T_L}$, with $T_L = 253$ K
$5 = \frac{253}{T_H - 253} \Rightarrow T_H = \frac{6}{5}\times 253 = 303.6$ K
$T_H \approx 31^\circ C$

Work done on the gas equals the area under the P–V curve; the adiabatic curve is the steepest, so it encloses the largest area.
$W_{isochoric} = 0$ and $W_{adiabatic} > W_{isothermal} > W_{isobaric}$

Isothermal: $P_1V_1 = P_2V_2 \Rightarrow PV = P_2(2V) \Rightarrow P_2 = \frac{P}{2}$
Adiabatic: $P_2V_2^\gamma = P_3V_3^\gamma \Rightarrow \frac{P}{2}(2V)^\gamma = P_3(16V)^\gamma$
$P_3 = \frac{P}{2}\left(\frac{1}{8}\right)^{5/3} = \frac{P}{64}$

Work done in the cycle = area enclosed by the P–V loops; the two triangular loops are traversed in opposite senses.
$W = \frac{1}{2}(2P_0 - P_0)(2V_0 - V_0) + \left[-\frac{1}{2}(3P_0 - 2P_0)(2V_0 - V_0)\right]$
$W = \frac{P_0V_0}{2} - \frac{P_0V_0}{2} = 0$

$P \propto T^3$ and $PV = nRT$ give $PV^{3/2}$ = constant.
Comparing with $PV^\gamma$ = constant: $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$

Net work done = area of triangle ABC
$= \frac{1}{2}\times[(7 - 2)\times10^{-3}][(6 - 2)\times10^5] = 1000$ J

Isothermal expansion: PV = constant, so as the volume goes from V to 3V the pressure falls along a rectangular hyperbola.
Then the volume is reduced from 3V back to V at constant pressure: a horizontal straight line directed towards smaller volume.
So the diagram shows a falling hyperbola from A (at V) to 3V, followed by a horizontal line back to V ending at B, which lies below A.

The cycle is a rectangle with pressures P and 2P and volumes V and 3V.
Area enclosed $= (2P - P)(3V - V) = 2PV$
The cycle is traversed anticlockwise, so the work done by the gas is negative: $W = -2PV$
For a complete cycle $\Delta E = 0$, so $Q = W + \Delta E = -2PV$
So the heat rejected by the gas is 2PV.

Internal energy is a state function. All three processes have the same initial state A and final state B, so $\Delta U_1 = \Delta U_2 = \Delta U_3$.
Work done = area under the P–V curve. Path 1 lies highest and path 3 lowest, so $W_1 > W_2 > W_3$.
First law: $Q = \Delta U + W$; with equal $\Delta U$, $Q_1 > Q_2 > Q_3$.

$T_1 = 27 + 273 = 300$ K; $T_2 = 927 + 273 = 1200$ K; $P_1 = 2$ atm.
For an adiabatic process $P^{1-\gamma}T^\gamma$ = constant, so $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma - 1}}$
$\frac{\gamma}{\gamma - 1} = \frac{1.4}{0.4} = \frac{7}{2}$
$P_2 = 2\times\left(\frac{1200}{300}\right)^{7/2} = 2\times(4)^{7/2}$
$(4)^{7/2} = 2^7 = 128$
$P_2 = 2\times128 = 256$ atm

Heat absorbed in melting: $Q = mL = 1000\ g\times80\ cal/g = 8\times10^4$ cal
Melting takes place at the constant temperature T = $0^\circ C$ = 273 K.
Change in entropy $\Delta S = \frac{Q}{T}$
$\Delta S = \frac{8\times10^4}{273}$
$\Delta S \approx 293$ cal/K
Note: the source prints the latent heat as '80 cal/°C'; the correct unit, cal/g, is used here.

First law of thermodynamics: $Q = \Delta U + W$, where W is the work done by the gas.
For an ideal gas the internal energy depends only on temperature; in an isothermal process $\Delta T = 0$, so $\Delta U = 0$.
Given W = −150 J.
$Q = 0 + (-150) = -150$ J
The negative sign means heat has left the gas.
Hence 150 J of heat has been removed from the gas.

For an adiabatic process: $PV^\gamma$ = constant; for a monoatomic gas $\gamma = \frac{5}{3}$
$P_1V_1^{5/3} = P_2\left(\frac{V_1}{8}\right)^{5/3}$
$P_2 = P_1(8)^{5/3}$
$(8)^{5/3} = (2^3)^{5/3} = 2^5 = 32$
$P_2 = 32P_1$

First law of thermodynamics: $\Delta Q = \Delta U + \Delta W$
In an adiabatic process no heat is exchanged: $\Delta Q = 0$
$0 = \Delta U + \Delta W$, so $\Delta U = -\Delta W$
In an isothermal process of an ideal gas the temperature is constant, so $\Delta U = 0$ and $\Delta Q = \Delta W$; neither isothermal relation given is true.
Hence $\Delta U = -\Delta W$ in an adiabatic process.

Adiabatic process: no heat is exchanged, so the system is insulated from the surroundings, and for an ideal gas $PV^\gamma$ = constant. Both statements are true.
Isothermal process: the temperature remains constant (true).
Isochoric process: it is the volume that remains constant; the pressure changes with temperature ($P \propto T$).
The process in which pressure remains constant is the isobaric process.
Hence the statement that is not true is 'In an isochoric process pressure remains constant'.

Heat absorbed: $Q = 2$ kcal $= 2\times1000\times4.2$ J $= 8400$ J
Work done by the system: W = 500 J
First law of thermodynamics: $Q = \Delta U + W$
$\Delta U = Q - W = 8400 - 500$
$\Delta U = 7900$ J

In a cyclic process the system returns to its initial state, so the initial and final states are the same.
Internal energy is a state function; it depends only on the state of the system.
Therefore the initial and final internal energies are equal and the change in internal energy is zero: $E = 0$.
From the first law, $Q = E + W$, so for a cycle $Q = W$.
$Q$ and $W$ are not zero in general (the work equals the area enclosed by the cycle on the P-V diagram); they are only equal to each other.
Hence the correct relation is $E = 0$.

Efficiency of the engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the temperature of the source and $T_2$ that of the sink (in kelvin).
According to the problem, $\dfrac{1}{6} = 1 - \dfrac{T_2}{T_1}$ ...(1)
From (1): $\dfrac{T_2}{T_1} = \dfrac{5}{6}$
When the temperature of the sink is reduced by $62^\circ C$ (a fall of 62 K), the efficiency is doubled: $2\left(\dfrac{1}{6}\right) = 1 - \dfrac{T_2 - 62}{T_1}$ ...(2)
$\dfrac{1}{3} = 1 - \dfrac{T_2}{T_1} + \dfrac{62}{T_1} = \dfrac{1}{6} + \dfrac{62}{T_1}$
$\dfrac{62}{T_1} = \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{6}$
$T_1 = 62 \times 6 = 372$ K
$T_1 = 372 - 273 = 99^\circ C$ = temperature of the source.
(The sink is at $T_2 = \dfrac{5}{6} \times 372 = 310$ K $= 37^\circ C$.)

Efficiency of a Carnot engine: $\eta = 1 - \dfrac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ the sink temperature.
$1 - \dfrac{300}{T_1} = 0.4 \Rightarrow \dfrac{300}{T_1} = 0.6 \Rightarrow T_1 = 500$ K
The efficiency is to be increased by 50 percent of the original efficiency: $\eta' = 0.4 + 0.4 \times \dfrac{50}{100} = 0.6$
Let the source temperature be increased by $\Delta T$, the sink remaining at 300 K.
$0.6 = 1 - \dfrac{300}{500 + \Delta T}$
$\dfrac{300}{500 + \Delta T} = 0.4 \Rightarrow 500 + \Delta T = 750$
$\Rightarrow \Delta T = 250$ K

A reversible process is one that can be retraced in the opposite direction, so that the system and the surroundings pass through exactly the same states in reverse order.
It must be carried out very slowly (quasi-statically), with the system always in equilibrium, and without dissipative effects such as friction, resistance or heat flow across a finite temperature difference.
A slow isothermal compression of a gas, in contact with a heat reservoir at the same temperature, satisfies these conditions; it can be reversed by an equally slow isothermal expansion. So it is a reversible process.
Transfer of heat by conduction or radiation takes place from a hotter to a colder body across a finite temperature difference; heat never flows back by itself, so these are irreversible.
Electrical (Joule) heating of a nichrome wire is a dissipative process; the heat produced cannot be converted back into electrical energy in the same way, so it is irreversible.
So the reversible process is isothermal compression.

$T_1 = 227 + 273 = 500$ K (source) and $T_2 = 127 + 273 = 400$ K (sink)
Efficiency: $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{T_2}{T_1}$
$\eta = 1 - \dfrac{400}{500} = 0.2$
$W = Q_1\left(1 - \dfrac{T_2}{T_1}\right) = 6 \times 10^{4} \times 0.2$
$W = 1.2 \times 10^{4}$ cals

In an adiabatic process no heat is exchanged: $Q = 0$, so by the first law $0 = \Delta U + W \Rightarrow \Delta U = -W$.
The gas does work $W = 6R$, so $\Delta U = -6R$.
For an ideal gas, $\Delta U = \mu C_V \Delta T$, with $C_V = \dfrac{R}{\gamma - 1}$.
Here $\gamma = \dfrac{5}{3}$, so $C_V = \dfrac{R}{\frac{5}{3} - 1} = \dfrac{3}{2}R$
Therefore $-6R = 1 \times \dfrac{3}{2}R\,\Delta T$
$\Rightarrow \Delta T = -4$ K
$T_{final} = (T - 4)$ K (the gas cools as it does work at the cost of its internal energy).