(a) Let capacity of each cylinder be V and atmospheric pressure be p.
p₁ = p
Initial volume of gas = Volume of cylinder A
V₁ = V
When stopcock is opened, then volume available for gas becomes 2V
V₂ = 2V
Final pressure (p₂) = ?
As system is thermally insulated, therefore there is no change in temperature during the process and hence it is an isothermal process.
For an isothermal process (according to Boyle's law),
p₁V₁ = p₂ V₂
or    \({ p }_{ 2 }={ p }_{ 1 }\frac { { V }_{ 1 } }{ { V }_{ 2 } } =pX\frac { V }{ 2V }\)
\( \\ =\frac { p }{ 2 } =\frac { 1 }{ 2 } atm=0.5\ atm\)
(b) Change in internal energy, \(\triangle \)U = 0, as work is done on or by the gas.
(c) Change in temperature of the gas is zero as gas does no work in expansion.
(d) No, because free expansion of gas is rapid and cannot be controlled. The intermediate states are non-equilibrium states and so not satisfy the gas equation. Therefore, the intermediate state of the gas does not be on the p - V - T surface.

Given, T₁ = 27⁰C = 27 + 273 = 300K
T₂ = 0⁰ C = 0 + 273
= 273K
As we know, to freeze one gram of water 0⁰C, 80cal of heat must be transferred from water at 0^oC to the surrounding at 27⁰C.
Q₂ = 80 cal
The coefficient of performance of a refrigerator,
\(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \Rightarrow \frac { 80 }{ W } =\frac { 273 }{ 300-273 } \)
\(\\ \Rightarrow 7.91\ cal\) 

\(\triangle U=\triangle Q-\triangle W \left[ from\ first\ law \right]\)
\( \\ =70-20=50\ cal\)

According to the second law of thermodynamics whole of the heat cannot be converted completely into work. Some part of the heat that is not converted into work is released by the engine to the atmosphere /9as sink).
Thus, thermal pollution will always occur.

Given, T₂ = 9⁰C = 9 + 273 = 282K
T₁ = 36⁰C = 36 + 273 = 309K
Coefficient of performance
\(=\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } =\frac { 282 }{ 309-282 } =10.4\) 

As we know,
\(\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 3 }{ 5 } \)
\(\\ \because \ 1-\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
\(\\ \Rightarrow \frac { { Q }_{ 1 }-{ Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { 500-300 }{ 500 } \)
\(\\ \Rightarrow \frac { W }{ { Q }_{ 1 } } =\frac { 2 }{ 5 } \)
\(\\ \therefore \ { Q }_{ 1 }={ 10 }^{ 3 }\times \frac { 5 }{ 2 } =2500J\)
and heat transferred by the engine to the cold reservoir in one cycle
Q₂ = Q₁ - W = 2.5 kJ- 1 kJ = 1.5 k J

The efficiency of a heat engine is stated as the ratio of the network engine and heat absorbed by the working substance.
Suppose a heat engine absorbs Q₁ heat from the hot reservoir and gives Q₂ heat rejected to the colder reservoir. So, the work done by the working substance is
 W = Q₁ - Q₂
So, efficiency of heat engine
 \(\eta =\frac { W }{ { Q }_{ 1 } } =\frac { { Q }_{ 1 }-{ Q }_{ 2 } }{ { Q }_{ 1 } } =1-\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } \)

The efficiency of a carnot engine depends, on the temperature of source of heat and the sink.

A reversible process is never possible in nature because of dissipative forces and condition for a quasi-static process is not practically possible.

Heat rejected by refrigerator remains in the room itself and so, temperature of room increases.Hence, it is false.

Leaking of air is adiabatic expansion and adiabatic expansion produces cooling.

If two isothermal curves intersect, this implies that the pressure and volume of a gas are the same at two different temperatures, that's impossible.

Let the first engine take Q₁ heat as input at temperature,
T₁ = 800 K
and gives out heat Q₂ at temperature T₀ The second engine receive Q₂ as input and give is out heat Q₃ at temperature T₃ = 300 K to the sink.
Work done by first (A) engine = work done by second (B) engine.
Thus, Q₁ - Q₂ = Q₂ - Q₃
Dividing both sides by Q₁
\(1-\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } -\frac { { Q }_{ 3 } }{ { Q }_{ 1 } } \)
\(\\ \Rightarrow 1-T/{ T }_{ 1 }=\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } (1-\frac { { Q }_{ 3 } }{ { Q }_{ 2 } } )\)
\(\\ \Rightarrow -T/{ T }_{ 1 }=\frac { T }{ { T }_{ 1 } } (1-{ T }_{ 3 }/T)\)
\(\\ \Rightarrow { T }_{ 1 }/T-1=1-\frac { { T }_{ 3 } }{ T }\)
\( \Rightarrow \frac { { T }_{ 1 } }{ T } +\frac { { T }_{ 3 } }{ T } =2\)
\(\\ \Rightarrow \frac { 1 }{ T } ({ T }_{ 1 }+{ T }_{ 3 })=2\)
\(\Rightarrow T=\frac { { T }_{ 1 }+{ T }_{ 3 } }{ 2 } \)
\(\\ \Rightarrow T=\frac { 900+400 }{ 2 } =650K\)   ​​​​​​

Here, Q₂ = 250 Js^-1
T₂ = -23^o C = -23 + 273 = 250 K
T₁ = 25^o C = 25 + 273 = 298 K
We know, \(\beta =\frac { { Q }_{ 2 } }{ W } =\frac { { T }_{ 2 } }{ { T }_{ 1 }-{ T }_{ 2 } } \)
\(W=\frac { { Q }_{ 2 }({ T }_{ 1 }-{ T }_{ 2 }) }{ { T }_{ 2 } } \)
\(\\ W=\frac { 250(298-250) }{ 250 } =\frac { 250\times48 }{ 250 } \)
\(\\ W=48{ Js }^{ -1 }\)

A heat engine is a device (or a combination) which converts heat into work.
Its efficiency, \(\eta =\frac { Work\ output }{ Heat\ input } \)
\(\eta =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
Where, T₂  = temperature of sink
T₁  = temperature of source.
From above expression, we can see that for 100% efficiency, T₂ = 0
It is impossible to design a thermal engine that has 100% efficiency because it is not possible to have a sink with kelvin temperature.

Let the initial temperatures of hot and cold reservoirs were T₁ and T₂ . The efficiency of the Carnot engine is given by
So, initially
\(\eta =\frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } ...(i)\)
As given the temperature of both the reservoirs is raised by equal amount t , so T₁^' = T₁ + t and   T ₂^' = T + t ₂ + t . The final efficient of the Carnot engine will be
\({ \eta }^{ ' }=\frac { { T }_{ 1 }-{ T }_{ 2 }^{ ' } }{ { T }_{ 1 }^{ ' } } =\frac { ({ T }_{ 1 }+t)-({ T }_{ 2 }+t) }{ ({ T }_{ 1 }+t) } \)
= \(\frac { { T }_{ 1 }-{ T }_{ 2 } }{ ({ T }_{ 1 }+t) } ...(ii)\)
Dividing Eq. (ii) by Eq. (i), we have
\(\frac { { \eta }^{ ' } }{ \eta } =\frac { \left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 }+t } \right) }{ \left( \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right) } =\frac { { T }_{ 1 } }{ { T }_{ 1 }+t } ...(iii)\)
As \({ \eta }^{ ' }<\eta \)  i.e., the efficiency of Carnot engine decreases.

Given, work done (W) = - 22.3 J
Work done is taken negative as work is done on the system.
In an adiabatic change, \(\Delta \)Q = 0
Using first law of thermodynamics
\(\Delta \).U = \(\Delta \)Q - W = 0 -(- 22.3) = 22.3J
For another process between states A and B,
Heat absorbed (\(\Delta \)Q) = + 9.35 cal
= + (9.35 x 4.19) J = + 39.18 J
Change in internal energy between two states via different paths are equal.
\(\because \)   \(\Delta \)U = 22.3 J
\(\therefore \)   From first law of thermodynamics,
\(\Delta \)U = \(\Delta \)Q - W
or W = \(\Delta \)Q -\(\Delta \)U
= 39.18 - 22.3
= 16.88J \(\approx \)16.9J