181.
An ideal gas is compressed to half its initial volume by means of several process. Which of the process results in the maximum work done on the gas?
Given, ideal gas is compressed to half its initial volume i.e.
$${V_0} = \frac{V}{2}$$
The isochoric process is one in which volume is kept constant, meaning that work done by the system will be zero. $${W_{{\text{isochoric}}}} = 0$$
As we know, work done on the gas = Area under curve, i.e.
$${W_{{\text{adiabatic}}}} > {W_{{\text{isothermal}}}} > {W_{{\text{isobaric}}}}$$
182.
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $${V_1}$$ and contains ideal gas at pressure $${P_1}$$ and temperature $${T_1}.$$ The other chamber has volume $${V_2}$$ and contains ideal gas at pressure $${P_2}$$ and temperature $${T_2}.$$ If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be
A
$$\frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$
B
$$\frac{{{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}}}{{{P_1}{V_1} + {P_2}{V_2}}}$$
C
$$\frac{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}{{{P_1}{V_1} + {P_2}{V_2}}}$$
D
$$\frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}}}$$
Here $$Q = 0$$ and $$W = 0.$$ Therefore from first law of thermodynamics $$\Delta U = Q + W = 0$$
$$\therefore $$ Internal energy of the system with partition = Internal energy of the system without partition.
$$\eqalign{
& {n_1}{C_v}{T_1} + {n_2}{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}T \cr
& \therefore T = \frac{{{n_1}{T_1} + {n_2}{T_2}}}{{{n_1} + {n_2}}} \cr
& {\text{But}}\,{n_1} = \frac{{{P_1}{V_1}}}{{R{T_1}}}\,{\text{and}}\,{n_2} = \frac{{{P_2}{V_2}}}{{R{T_2}}} \cr
& \therefore T = \frac{{\frac{{{P_1}{V_1}}}{{R{T_1}}} \times {T_1} + \frac{{{P_2}{V_2}}}{{R{T_2}}} \times {T_2}}}{{\frac{{{P_1}{V_1}}}{{R{T_1}}} + \frac{{{P_2}{V_2}}}{{R{T_2}}}}} = \frac{{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)}}{{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}} \cr} $$
183.
An ideal gas has temperature $${T_1}$$ at the initial state i shown in the $$P-V$$ diagram. The gas has a higher temperature $${T_2}$$ at the final states $$a$$ and $$b,$$ which it can reach the paths shown. The change in entropy:
$$\eqalign{
& {Q_a} = \Delta U + 0 = \Delta U\,\,{\text{and}}\,\,{Q_b} = \Delta U + P\Delta V \cr
& {\text{As}}\,\,{Q_b} > {Q_a} \cr} $$
∴ Change in entropy is greater in case (b).
184.
An ideal gas undergoing adiabatic change has the following pressure-temperature relationship
A
$${p^{\gamma - 1}}{T^\gamma } = {\text{constant}}$$
B
$${p^\gamma }{T^{\gamma - 1}} = {\text{constant}}$$
C
$${p^\gamma }{T^{1 - \gamma }} = {\text{constant}}$$
D
$${p^{1 - \gamma }}{T^\gamma } = {\text{constant}}$$
The adiabatic relation between $$p$$ and $$V$$ for a perfect gas is $$p{V^\gamma } = k\,......\left( {\text{i}} \right)$$
From standard gas equation $$pV = RT\,\,{\text{or}}\,\,V = \frac{{RT}}{p}$$
Putting value of $$V$$ in Eq. (i)
$$\eqalign{
& p{\left( {\frac{{RT}}{p}} \right)^\gamma } = k \cr
& {\text{or}}\,\,{p^{1 - \gamma }}{T^\gamma } = \frac{k}{{{R^\gamma }}} = {\text{another constant}} \cr
& {\text{i}}{\text{.e}}{\text{.,}}\,\,{p^{1 - \gamma }}{T^\gamma } = {\text{constant}} \cr} $$
185.
The slopes of isothermal and adiabatic curves are related as
A
isothermal curve slope $$=$$ adiabatic curve slope
$$\frac{{{\text{Slope}}\,{\text{of}}\,{\text{adiabatic}}\,{\text{curve}}}}{{{\text{Slope}}\,{\text{of}}\,{\text{isothermal}}\,{\text{curve}}}} = \frac{{{{\left( {\frac{{dP}}{{dV}}} \right)}_{{\text{adi}}}}}}{{{{\left( {\frac{{dP}}{{dV}}} \right)}_{{\text{iso}}}}}} = + \gamma $$
So slope to adiabatic curve is $$\gamma \left( { = \frac{{{C_p}}}{{{C_v}}}} \right)$$ times of isothermal curve, as clear also from figure.
186.
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $$\frac{{{C_P}}}{{{C_V}}}$$ for the gas is
According to question,
$$p \propto {T^3}\,......\left( {\text{i}} \right)\left( {_{T = {\text{temperature}}}^{p\, = \,{\text{pressure}}}} \right)$$
and we know that
$$pV = nRT\,\,{\text{and}}\,\,pV \propto T\,......\left( {{\text{ii}}} \right)$$
So, putting Eq. (ii) in (i),
$$\eqalign{
& p \propto {\left( {pV} \right)^3} \cr
& \Rightarrow {p^2}{V^3} = {\text{constant}} \cr
& \Rightarrow p{V^{\frac{3}{2}}} = {\text{constant}}\,......\left( {{\text{iii}}} \right) \cr} $$
$$ \Rightarrow $$ Comparing Eq. (iii) with $$pV\gamma = {\text{constant}}{\text{.}}$$
We have $$\gamma = \frac{3}{2}.$$
187.
A diatomic ideal gas is used in a car engine as the working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from $$V$$ to $$32\,V,$$ the efficiency of the engine is
188.
An ideal gas heat engine operates in Carnot cycle between $${227^ \circ }C$$ and $${127^ \circ }C.$$ It absorbs $$6 \times {10^4}cal$$ of heat at higher temperature. Amount of heat converted to work is
190.
$$1\,gm$$ of water at a pressure of $$1.01 \times {10^5}\,Pa$$ is converted into steam without any change of temperature. The volume of $$1\,g$$ of steam is $$1671\,cc$$ and the latent heat of evaporation is $$540\,cal.$$ The change in internal energy due to evaporation of $$1\,gm$$ of water is