Heat and Thermodynamics
One mole of a monatomic gas is taken through a cycle ABCDA as shown in the PV diagram. Column II give the characteristics involved in the cycle. Match them with each of the processes given in Column I.
| Column I | Column II | ||
|---|---|---|---|
| (A) | Process A$ \to $ B | (P) | Internal energy decreases. |
| (B) | Process B$ \to $C | (Q) | Internal energy increase. |
| (C) | Process C$ \to $D | (R) | Heat is lost. |
| (D) | Process D$ \to $A | (S) | Heat is gained. |
| (T) | Work is done on the gas. |
One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P0. Choose the correct option(s) from the following:
Explanation:
The diatomic gas has five degrees of freedom i.e., f = 5. Thus, the internal energy per mole, specific heat at constant volume, specific heat at constant pressure, and the ratio of specific heats for a diatomic gas, are given by
$U = (f/2)RT = (5/2)RT$, ${C_v} = dU/dT = 5R/2$, ${C_p} = {C_v} + R = 7R/2$, $\gamma = {C_p}/{C_v} = 7/5$.
For an adiabatic process, $T{V^{\gamma - 1}}$ = constant
${T_i}V_i^{\gamma - 1} = {T_f}V_f^{\gamma - 1}$
Substituting the given values, we get
${T_i}V_i^{\gamma - 1} = a{T_i}{\left( {{{{V_i}} \over {32}}} \right)^{\gamma - 1}} \Rightarrow a = {32^{\gamma - 1}}$
For diatomic gas, $\gamma = {7 \over 5}$
$a = {32^{{7 \over 5} - 1}} = {32^{2/5}} = {2^2} = 4$
Explanation:
In the final state, ice-water mixture is in equilibrium. Thus, the temperature of m grams of ice is raised from $-$5 $^\circ$C to 0 $^\circ$C. The heat absorbed in this process is
Q1 = mS$\Delta$T. ...... (1)
The state of m1 = 1 g of ice is changed from solid to liquid. The heat absorbed in the melting process is
Q2 = m1L. ........ (2)
The heat supplied is Q = 420 J. By energy conservation, Q = Q1 + Q2. Substitute Q1 and Q2 from equations (1) and (2) to get
$m = {{Q - {m_1}L} \over {S\Delta T}} = {{420 - ({{10}^{ - 3}})(3.36 \times {{10}^5})} \over {(2100)(5)}}$
$ = 8 \times {10^{ - 3}}$ kg = 8 g.
Explanation:
According to Wien's displacement law, $\lambda$mT = constant
$\therefore$ ${({\lambda _m})_A}{T_A} = {({\lambda _m})_B}{T_B}$
or, ${{{T_A}} \over {{T_B}}} = {{{{({\lambda _m})}_B}} \over {{{({\lambda _m})}_A}}} = {{1500\,nm} \over {500\,nm}}$ or ${{{T_A}} \over {{T_B}}} = 3$ ...... (i)
According to Stefan Boltzmann law, rate of energy radiated by a black body
$E = \sigma A{T^4} = \sigma 4\pi {R^2}{T^4}$ [Here, $A = 4\pi {R^2}$]
$\therefore$ ${{{E_A}} \over {{E_B}}} = {\left( {{{{R_A}} \over {{R_B}}}} \right)^2}{\left( {{{{T_A}} \over {{T_B}}}} \right)^4} = {\left( {{{6\,cm} \over {18\,cm}}} \right)^2}{(3)^4}$ (Using (i))
$ = 9$
Statement - 2: $R = {R_0}\left( {1 + \alpha \,\Delta t} \right)$ is valid only when the change in the temperature $\Delta T$ is small and $\Delta T = \left( {R - {R_0}} \right) < < {R_0}.$
Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is :
The work done on the gas in taking it from $D$ to $A$ is :
The net work done on the gas in the cycle $ABCDA$ is:
The figure shows the PV plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semicircle and CDA is half of an ellipse. Then,
$C_V$ and $C_P$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then
A metal rod AB of length 10x has its one end A in ice at 0$^\circ$C and the other end B in water at 100$^\circ$C. If a point P on the rod is maintained at 400$^\circ$C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of $\lambda x$ from the ice end A, find the value of $\lambda$. (Neglect any heat loss to the surrounding.)
Explanation:
The steady rate of flow of heat is
${{\Delta Q} \over {\Delta t}} = kA\left( {{{\Delta T} \over {\Delta x}}} \right)$
Therefore, ${{kA(400 - 0^\circ )} \over {\lambda x}} = {m_{ice}}\,{l_{ice}}$
${{kA(400 - 100)} \over {(10 - \lambda )x}} = {m_{water}}\,{l_{water}}$
where ${m_{ice}}$ is the mass of ice melted per unit times and ${m_{water}}$ is the mass of water evaporated per unit times. It is given that
${m_{ice}} = {m_{water}}$
Therefore, ${{kA \times 400} \over {\lambda x({l_{ice}})}} = {{kA \times 300} \over {(10 - \lambda )x\,{l_{water}}}}$
$ \Rightarrow 4(10 - \lambda ) \times 540 = 3\lambda \times 80$
$ \Rightarrow 5400 - 540\lambda = 60\lambda $
$ \Rightarrow \lambda = 9$
Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II:
| Column I | Column II | ||
|---|---|---|---|
| (A) | The energy of the system is increased. | (P) | System : A capacitor, initially uncharged. Process : It is connected to a battery. |
| (B) | Mechanical energy is provided to the system, which is converted into energy of random motion of its parts. | (Q) | System : A gas in an adiabatic container filled with an adiabatic piston. Process : The gas is compressed by pushing the piston. |
| (C) | Internal energy of the system is converted into its mechanical energy. | (R) | System : A gas in a rigid container. Process : The gas gets cooled due to colder atmosphere surrounding it. |
| (D) | Mass of the system is decreased. | (S) | System : A heavy nucleus, initially at rest. Process : The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted. |
| (T) | System : A resistive wire loop. Process : The loop is placed in a time varying magnetic field perpendicular to its plane. |
Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) | An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened. |
(P) | The temperature of the gas decreases |
| (B) | An ideal monatomic gas expands to twice its original volume such that its pressure P $\propto$ $\frac{1}{\mathrm{V}^2}$, where V is the volume of the gas | (Q) | The temperature of the gas increase or remains constant. |
| (C) | An ideal monoatomic gas expands to twice its original volume such that its pressure P $\propto$ $\frac{1}{\mathrm{V}^{4/3}}$, where V is its volume | (R) | The gas loses heat |
| (D) | An ideal monoatomic gas expands such that its pressure P and volume V follows the behaviour shown in the graph |
(S) | The gas gains heat |
An ideal gas is expanding such that PT$^2$ = constant. The coefficient of volume expansion of the gas is
STATEMENT 1
The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume.
Because
STATEMENT 2
The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.
The piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height H of the water column in the cylinder satisfies
Where r0 is the radius of the Earth and $\sigma $ is Stefan's constant.
In a dark room with ambient temperature $\mathrm{T}_0$, a black body is kept at a temperature T . Keeping the temperature of the black body constant (at T), sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is/are correct?
The quantity of radiation absorbed by the black body in unit time will increase.
Since emissivity $=$ absorptivity, hence the quantity of radiation emitted by black body in unit time will increase.
Black body radiates more energy in unit time in the visible spectrum.
The reflected energy in unit time by the black body remains the same.
In an insulated vessel, 0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed. Then, find the final temperature of the mixture.
Explanation:
$ 0.05 \mathrm{~kg} \text { steam at } 373 \mathrm{~K} \xrightarrow{\mathrm{Q}_1}0.05 \mathrm{~kg} \text { water at } 373 \mathrm{~K} \text {. } $
$ 0.05 \mathrm{~kg} \text { water at } 373 \mathrm{~K} \xrightarrow{\mathrm{Q}_2} 0.05 \mathrm{~kg} \text { water at } 273 \mathrm{~K}$
$ 0.45 \mathrm{~kg} \text { ice at } 253 \mathrm{~K} \xrightarrow{\mathrm{Q}_3} 0.45 \mathrm{~kg} \text { ice at } 273 \mathrm{~K} $
$ 0.45 \mathrm{~kg} \text { ice at } 273 \mathrm{~K} \xrightarrow{\mathrm{Q}_4} 0.45 \mathrm{~kg} \text { water at } 273 \mathrm{~K}$
$ \begin{aligned} &\begin{aligned} & \Rightarrow \mathrm{Q}_1=m \mathrm{~L}=50(540)=27000 \mathrm{cal}=27 \mathrm{kcal} (\mathrm{Q}=m \mathrm{~L}) \\ & \mathrm{Q}_2=m s \Delta t=(50)(1)(100)=500 \mathrm{cal}=5 \mathrm{kcal} \\ & \mathrm{Q}_3=m s \Delta t=(4.50)(0.5)(20)=4500 \mathrm{cal}=4.5 \mathrm{kcal} \\ & \mathrm{Q}_4=(m \mathrm{~L})=450 \times 580=36000 \mathrm{cal}=36 \mathrm{kcal} \,\, [fusion]\end{aligned}\\ &\text { } \end{aligned} $
Since, $Q_1+Q_2>Q_3$ but $Q_1+Q_2 Therefore temperature of mixture $=273 \mathrm{~K}$ (or $0^{\circ} \mathrm{C}$ ) Method II Heat lost by steam to covert into $0^{\circ} \mathrm{C}$ water $ \begin{gathered} \mathrm{H}_{\mathrm{L}}=0.05 \times 540+0.05 \times 10 \times 1 \\ {\left[\mathrm{Q}=\mathrm{H}_{\mathrm{L}}=m \mathrm{~L}+m s \Delta t\right]} \\ \mathrm{H}_{\mathrm{L}}=27+5=32 \mathrm{kcal} \end{gathered} $ Heat required by ice to change into $0^{\circ} \mathrm{C}$ water. $ \begin{aligned} \mathrm{H}_{\mathrm{g}} & =450(20) \frac{1}{2}+450 \times 80 \\ & =4500+36000=40500 \mathrm{cal} \\ & =40.5 \mathrm{kcal} \end{aligned} $ $\Rightarrow$ To convert ice, we have $450 \times \frac{1}{2}=225 \mathrm{cal}$. But we need 32000 cal. Therefore, ice will remain as ice. Hence final temperature is $0^{\circ} \mathrm{C}$.
Heat given to the processes is positive. Match Column I with Column II:
| Column I | Column II | ||
|---|---|---|---|
| (A) | JK | (P) | $ \Delta W>0 $ |
| (B) | KL | (Q) | $ \Delta \mathrm{Q}<0 $ |
| (C) | LM | (R) | $ \Delta \mathrm{W}<0 $ |
| (D) | MJ | (S) | $ \Delta Q>0 $ |
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{~S}) ; \mathrm{C} \rightarrow(\mathrm{Q}) ; \mathrm{D} \rightarrow( \mathrm{R})]$
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}) ; \mathrm{C} \rightarrow(\mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{Q})]$
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}, \mathrm{~S}) ; \mathrm{C} \rightarrow(\mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{Q}, \mathrm{R})] $
$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{S}) ; \mathrm{C} \rightarrow(\mathrm{S}) ; \mathrm{D} \rightarrow( \mathrm{R})]$
A cylinder of mass $1 \mathrm{~kg}$ is given heat of $20000 \mathrm{~J}$ at atmospheric pressure. If initially temperature of cylinder is $20^{\circ} \mathrm{C}$, find
(A) The final temperature of the cylinder;
(B) The work done by the cylinder;
(C) The change in internal energy of the cylinder.
Given :
The specific heat of cylinder
$=400 \mathrm{~J} \mathrm{~kg}^{-1 \circ} \mathrm{C}^{-1}$
Coefficient of volume expansion
$=9 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} \text {; }$
Atmospheric pressure $=10^{5} \mathrm{~N} / \mathrm{m}^{2}$ Density of cylinder $=9000 \mathrm{~kg} / \mathrm{m}^{3}$ )
