As no heat is allowed to be exchanged, the process is adiabatic.
\(\therefore \) \({ \rho }_{ 2 }V_{ 1 }^{ y }\quad ={ \rho }_{ 1 }V_{ 1 }\quad or\quad \frac { { p }_{ 2 } }{ { p }_{ 1 } } =\left( \frac { { V }_{ 1 } }{ { V }_{ 2 } } \right) ^{ y }\)
As \({ V }_{ 2 }=\frac { 1 }{ 2 } { V }_{ 1 }or\frac { { p }_{ 2 } }{ { p }_{ 1 } } =\left( \frac { { V }_{ 1 } }{ { 1 }/{ 2{ V }_{ 1 } } } \right) ^{ 1.4 }={ 2 }^{ 1.4 }=2.64\)

Here, volume of water heated = 3.0 L/min
Mass of water heated, m = 3000 g/min
Rise in temperature, \(\Delta \)T = 77 - 27 =50^oC
Specific heat of water, C = 4.2 J g^-1 oC^-1
Amount of heat used,\(\Delta \)Q = \(mC\Delta T=3000\times 4.2\times 50\)
\(\\ =63\times 10^{ 4 }\ J/min\)
Heat of combination = 4 \(\times \)10⁴ J/g
Rate of combustion of fuel = \(\frac { 63\times { 10 }^{ 4 } }{ 4\times { 10 }^{ 4 } } \) 
= 15.75g / min

Here, mass of gas, m =  2\(\times \)10^-2 kg = 20g
Rise in temperature, \(\Delta \)T = 45^oC
Heat required,\(\Delta \)Q = ?
Molecular mass, M =28
Number of moles, n =\(\frac { m }{ n } =\frac { 20 }{ 28 } =0.714\)
As nitrogen is a diatomic gas, molar specific heat at constant pressure is
\(C_{ p }=\frac { 7 }{ 2 } R=\frac { 7 }{ 2 } \times 8.3J\quad mol^{ -1 }K^{ -1 }\)
\(\\ As\ \Delta Q=nC_{ p }\Delta T\)
\(\\ \therefore \ \Delta Q=0.714\times \frac { 7 }{ 2 } \times 8.3\times 45J=933.4\ J\)

Heat energy supplied per second by the heater
\(\Delta \)Q = 100 W = 100 J/s
Work done by the system ( \(\Delta \)  W) = + 75 Jls
Rate of change in internal energy (\(\Delta \)U) = ?
According to first law of thermodynamics,
\(\Delta \)U = \(\Delta \)Q - \(\Delta \)W
= 100 - 75 = 25 J/s
= 25W

As is clear from Fig.
Change in pressure, Δp = EF = 5.0 - 2.0 = 3.0 atm = 3.0 x 10⁵ Nm^-2
Change in volume, ΔV=DF = 600 - 300 = 300 cc = 300 x 10^-6 m³
Work done by the gas from 0 to E to F = area of ΔDEF
\(W={1\over 2}\times DF\times EF\)
\(={1\over 2}\times (300 \times 10^{-6})x (3.0 \times 10^5) = 45 J\)

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

(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂)/2 only when the thermal capacities of both the bodies are equal.
(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.
(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, the temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.
(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

(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.