(4)

As AB is isobaric process, work done $\left(W_{AB}\right)$$=3\times {10}^2(3-1)=600$$\text{J}$

(2)

When a thermo-dynamic system undergoes a change in such a way that there is no exchange of heat with the surrounding, then the process is known as adiabatic process and the slope of adiabatic is more than isothermal.

[IMAGE 108]

(2)

Since the entire system is thermally insulated, no heat flows into the system. When the stopcock is removed, the gas expands adiabatically.

(3)

Adiabatic process is the process in which no exchange of heat energy takes place between the gas and the surroundings,

i.e., $\Delta$Q = 0.

(1)

Given process is isobaric.

$\therefore dQ=n{C}_{p}dT \text{where }{C}_p\text{ is specific heat at constant pressure.}$

or   $dQ=n\left(\frac{{\displaystyle 5}}{{\displaystyle 2}}R\right)dT$

Also, $dW=PdV=nRdT$                             $(\because PV=nRT)$

Required ratio $=\frac{dW}{dQ}=\frac{nRdT}{n\left(\frac{5}{2}R\right)dT}=\frac{2}{5}$

(1)

In process I, volume is constant.

$\therefore$      Process I → Isochoric; P → C

As slope of curve II is more than the slope of curve III,

Process II → Adiabatic and Process III → Isothermal

$\therefore$     Q → A, R → D

In process IV, pressure is constant. Process IV → Isobaric; S → B

(4)

Process described by the equation,

$P{V}^3=\text{constant}$

For a polytropic process, $P{V}^{\alpha }=\text{constant}$

         $C=C_V+\frac{R}{1-\alpha }=\frac{3}{2}R+\frac{R}{1-3}=R$

(4)

[IMAGE 111]

$V_1=V, V_2=V/2$

On P–V diagram, Area under adiabatic curve > Area under isothermal curve.
So, compressing the gas through adiabatic process will require more work to be done.

(3)

[IMAGE 112]

The P-V diagram of an ideal gas compressed from its initial volume $V_0$ to $\frac{V_0}{2}$ by several processes is shown in the figure.

Work done on the gas = Area under P-V curve.

As area under the P-V curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.

(1)

[IMAGE 114]

As initial and final points for two paths are same so

$\Delta U_{ABC}=\Delta U_{AC}$

AB is isochoric process.

         $\Delta W_{AB}=0$

$∆Q_{AB}=\Delta U_{AB}=400$ $\text{J}$

BC is isobaric process.

$\Delta Q_{BC}=\Delta U_{BC}+\Delta W_{BC}$

$100=\Delta U_{BC}+6\times {10}^4(4\times {10}^{-3}-2\times {10}^{-3})$

$100=\Delta U_{BC}+12\times 10$

$\Delta U_{BC}=100-120=-20$$\text{J}$

As, $\Delta U_{ABC}=\Delta U_{AC}, \Delta U_{AB}+\Delta U_{BC}=\Delta Q_{AC}-\Delta W_{AC}$

$400-20=\Delta Q_{AC}-(2\times {10}^4\times 2\times {10}^{-3}+\frac{1}{2}\times 2\times {10}^{-3}\times 4\times {10}^4)$

$\Delta Q_{AC}=460$$\text{J}$