$$\eqalign{ & \eta = 1 - \frac{{{T_2}}}{{{T_1}}} \cr & {T_1} = T\,\,\left( {{\text{Temperature}}\,{\text{of}}\,{\text{hot}}\,{\text{reservoir}}} \right) \cr & {\text{For}}\,\eta = 40\% , \cr & \frac{{40}}{{100}} = 1 - \frac{{{T_2}}}{T} \Rightarrow \frac{{{T_2}}}{T} = \frac{3}{5} = {T_2} = \frac{3}{5}T \cr & {\text{For}}\,\eta = 50\% , \cr & \frac{{50}}{{100}} = 1 - \frac{{\frac{3}{5}T}}{{{T_1}}} \Rightarrow {T_1} = \frac{{6T}}{5} \cr} $$

$$W = P\left( {\Delta V} \right) = 1 \times {10^5} \times \left( {3.34 - 2 \times {{10}^{ - 3}}} \right) = 340 \times {10^3}\,J.$$

Slope of adiabatic curve $$ = \gamma \times {\text{slope of isothermal curve}}$$
$$ = 1.4 \times \left( { - 400} \right) = - 560\,Mpa/{m^3}$$

We know that $$\frac{{{V_a}}}{{{V_b}}} = \frac{{{V_d}}}{{{V_c}}} \Rightarrow \frac{{{V_a}}}{{{V_d}}} = \frac{{{V_b}}}{{{V_c}}}$$

Here $$T{V^{\gamma - 1}} = {\text{constant}}$$
As $$\gamma = \frac{5}{3},{\text{hence }}T{V^{\frac{2}{3}}} = {\text{constant}}$$
Now $${T_1}{L_1}^{\frac{2}{3}} = {T_2}{L_2}^{\frac{2}{3}}\,\,\left( {\because \,\,V \propto \,L} \right);$$
Hence, $$\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^{\frac{2}{3}}}$$

The thermodynamic state of a homogeneous system may be represented by certain specific thermodynamic variables such as pressure $$p,$$ volume $$V,$$ temperature $$T$$ and entropy $$S.$$ Out of these four variables, any two are independent and when they are known the others may be determined. Thus, there are only two independent variables and the others may be considered their functions. For complete knowledge of the system certain relations are required and for this purpose we introduce some functions of variables $$p,V,T$$  and $$S$$ known as thermodynamic functions. There are four principal thermodynamic functions
(i) Internal energy $$\left( U \right)$$
(ii) Helmholtz function$$\left( H \right)$$
(iii) Enthalpy $$\left( F \right)$$
(iv) Gibb’s energy $$\left( G \right)$$
Hence, work done is not thermodynamic function.

The entropy change of the body in the two cases is same as entropy is a state function.

$$\eqalign{ & \Delta Q = \Delta U + \Delta W \Rightarrow 2 \times {10^3} \times 4.2 = \Delta U + 500 \cr & \Rightarrow \Delta U = 7900\,J \cr} $$

When a enclosed gas is accelerated in the positive $$x$$ - direction then the pressure of the gas decreases along the positive $$x$$ - axis and follows the equation
$$\Delta P = - \rho \,a\,dx$$
where $$\rho $$ is the density and $$a$$ the acceleration of the container.
The result will be more pressure on the rear side and less pressure on the front side.

Initially
$${V_1} = 5.6\,\ell ,{T_1} = 273\,K,{P_1} = 1\,{\text{atm,}}$$
$$\gamma = \frac{5}{3}$$   (For monoatomic gas)
The number of moles of gas is $$n = \frac{{5.6\,\ell }}{{22.4\,\ell }} = \frac{1}{4}$$
Finally (after adiabatic compression)
$${V_2} = 0.7\,\ell $$
For adiabatic compression $${T_1}{V_1}^{\gamma - 1} = {T_2}{V_2}^{\gamma - 1}$$
$$\eqalign{ & \therefore \,\,{T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}} \cr & = {T_1}{\left( {\frac{{5.6}}{{0.7}}} \right)^{\frac{5}{3} - 1}} \cr & = {T_1}{\left( 8 \right)^{\frac{2}{3}}} \cr & = 4{T_1} \cr} $$
We know that work done in adiabatic process is
$$W = \frac{{nR\Delta T}}{{\gamma - 1}} = \frac{9}{8}\,R{T_1}$$