The steam engine is not a Carnot engine, because all the process involved in the steam engine are not perfectly reversible. But we can calculate the maximum possible efficiency of the steam engine by considering it as a Carnot engine.
\(\eta =1-\frac { { T }_{ L } }{ { T }_{ H } } =1-\frac { 300K }{ 523K } =0.43\)
The steam engine can have maximum possible 43% of efficiency, implying this steam engine can convert 43% of input heat into useful work and remaining 57% is ejected as heat. In practice the efficiency is even less than 43%.

When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = ms_v ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ

From the ideal gas equation, \(V=\left( \frac { \mu R }{ P } \right) T\)
V-T graph is a straight line passing the origin.
The slope = \(\frac { \mu R }{ P } \)
The slope of V-T graph is inversely proportional to the pressure. If the slope is greater, lower is the pressure.
Here P₁ has larger slope than P₂. So P₂ > P₁.

Here, the process is adiabatic compression. The volume is given and temperature is to be found. we can use the equation (8.38 )
\({ T }_{ i }{ V }_{ i }^{ \Upsilon -1 }{ =T }_{ f }{ V }_{ f }^{ \Upsilon -1 }.\)
T_i = 300 K (273 + 27°C = 300 K)
\({ V }_{ i }=V\& { V }_{ f }=\frac { V }{ 4 } \)
\({ T }_{ f }={ T }_{ i }{ \left( \frac { { V }_{ i } }{ { V }_{ f } } \right) }^{ \Upsilon -1 }\) = 300 K × 4^1.4-1 = 300K\(\times\)1.741
T2 ≈ 522 K or 2490C
This temperature is higher than the boiling point of water. So it is very dangerous to touch the nozzle of blocked pump when you pump air.

Consider a container of gas with volume V, pressure P and temperature T. If we add sand particles one by one slowly on the top of the piston, the piston will move inward very slowly. This can be taken as almost a quasi-static process. It is shown in the figure

Sand particles added slowly- quasi-static process

Work done by the system (body),
W = +500 kJ
Heat released from the system (body),
Q = –230 kJ
The change in internal energy of a body
= \(\Delta\)U= – 230 kJ – 500 kJ = – 730 kJ

Work done on the system (by the person while stirring), W = -30 kJ = -30,000J
Heat flowing out of the system,
Q = -5 kcal = 5\(\times\)4184 J =-20920 J
Using First law of thermodynamics
\(\Delta\)U = Q-W
\(\Delta\)U = -20,920 J - (-30,000) J
\(\Delta\)U = -20,920 J + 30,000 J = 9080 J
Here, the heat lost is less than the work done on the system, so the change in internal energy is positive.

The efficiency of heat engine is given by
\(\eta =1-\frac { { Q }_{ L } }{ { Q }_{ H } } \)
\(\eta =1-\frac { 300 }{ 500 } =1-\frac { 3 }{ 5 } \)
\(\eta \) = 1 – 0.6 = 0.4
The heat engine has 40% efficiency, implying that this heat engine converts only 40% of the input heat into work.

To determine the curve corresponding to higher temperature, draw a horizontal line parallel to x axis as shown in the figure. This is the constant pressure line. The volumes V₁ and V₂ belong to same pressure as the vertical lines from V₁ and V₂ meet the constant pressure line.

At constant pressure, higher the volume of the gas, higher will be the temperature. From the figure, as V₁ > V₂ we conclude T₁ > T₂. In general the isothermal curve closer to the origin, has lower temperature.

(a) We know that work done by the gas in an isothermal expansion
Since μ = 0.5
W = 0.5mol \(\times\)\(\frac { 8.31J }{ mol.K } \times 300 K\ In\ \left( \frac { 6L }{ 2L } \right) \)
W = 1.369 kJ
Note that W is positive since the work is done by the gas.
(b) From the First law of thermodynamics, in an isothermal process the heat supplied is spent to do work.Therefore, Q = W = 1.369 kJ. Thus Q is also positive which implies that heat flows into the system.
(c) For an isothermal process
P_iV_i = P_fV_f = μRT
\({ P }_{ f }=\frac { \mu RT }{ { V }_{ f } } =0.5mol\times \frac { 8.31J }{ mol.K } \times \frac { 300K }{ 6\times { 10 }^{ -3 }{ m }^{ 3 } } \)
= 207.75 k Pa