The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4 . It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10 . The hot reservoir of the heat pump is at a temperature of 300 K . Which of the following statements is/are correct :
Work extracted from the Carnot engine in one cycle is 60 J.
Temperature of the cold reservoir of the Carnot engine is 600 K.
Temperature of the cold reservoir of the heat pump is 270 K.
Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.
In the given $P-V$ diagram, a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is first compressed adiabatically from state $A$ to state $B$. Then it expands isothermally from state $B$ to state $C$. [Given: $\left(\frac{1}{3}\right)^{0.6} \simeq 0.5, \ln 2 \simeq 0.7$ ].
Which of the following statement(s) is(are) correct?
(Take Stefan-Boltzmann constant = 5.67 $ \times $ 10−8 Wm−2K−4 , Wien’s displacement constant = 2.90 $ \times $ 10−3 m-K, Planck’s constant = 6.63 $ \times $ 10−34 Js, speed of light in vacuum = 3.00 $ \times $ 108 ms−1)
in the range 3.15 $ \times $ 10−8 W to 3.25 $ \times $ 10−8 W
(Given, 21.2 = 2.3; 23.2 = 9.2; R is a gas constant)
Ignoring the friction between the piston and the cylinder, the correct statements is/are
The figure below shows the variation of specific heat capacity (C) of a solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is(are) correct to a reasonable approximation.
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:
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
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.
Explanation:
$ \begin{aligned} & \mathrm{nRT}_{\mathrm{w}}=\mathrm{P}_{\mathrm{w}} \mathrm{~V}_{\mathrm{w}}=1 \mathrm{~J} \\ & \mathrm{P}_{\mathrm{W}}=\frac{1}{64} \times 10^6 \mathrm{~Pa} \end{aligned} $
For WX process
$ \begin{aligned} & P_X V_X^y=P_W V_W^y \\ & \Rightarrow P_X=P_W\left(\frac{V_W}{V_X}\right)^y \end{aligned} $
amount of heat absorbed in XY process
$ \begin{aligned} \mathrm{Q} & =\mathrm{nCP} \Delta \mathrm{~T}=\mathrm{n} \times \frac{5}{2} \mathrm{R} \times\left[\mathrm{T}_{\mathrm{Y}}-\mathrm{T}_{\mathrm{X}}\right] \quad\left[\text { For monoatomic gas } \mathrm{C}_{\mathrm{P}}=\frac{5 \mathrm{R}}{2}\right] \\ \mathrm{Q} & =\frac{5}{2}\left[\mathrm{nRT}_{\mathrm{Y}}-\mathrm{nRT}_{\mathrm{X}}\right] \\ & =\frac{5}{2}\left[\mathrm{P}_{\mathrm{Y}} \mathrm{~V}_{\mathrm{Y}}-\mathrm{P}_{\mathrm{X}} \mathrm{~V}_{\mathrm{X}}\right] \\ & =\frac{5}{2} \mathrm{P}_{\mathrm{X}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \quad\left[\because \mathrm{P}_{\mathrm{X}}=\mathrm{P}_{\mathrm{Y}} ; \text { Isobaric process }\right] \\ & =\frac{5}{2} \times \mathrm{P}_{\mathrm{W}} \times\left[\frac{\mathrm{V}_{\mathrm{W}}}{\mathrm{~V}_{\mathrm{X}}}\right]^{\mathrm{y}}\left[\mathrm{~V}_{\mathrm{Y}}-\mathrm{V}_{\mathrm{X}}\right] \end{aligned} $
Putting values :
Q = 1.6 Joule
Explanation:
Extension in spring
$ \begin{aligned} & \mathrm{x}=0.5 \mathrm{~L}-0.4 \mathrm{~L} \\ & =0.1 \mathrm{~L} \end{aligned} $
FBD of piston
$ \begin{aligned} & \mathrm{kx}+\mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A} \\ & \mathrm{P}_2 \mathrm{~A}=\mathrm{P}_1 \mathrm{~A}-\mathrm{kx} \\ & \mathrm{P}_2=\mathrm{P}_1-\frac{\mathrm{kL}}{\mathrm{~A}(10)} ........(i) \\ & \mathrm{P}_1 \mathrm{~V}=\mathrm{n}_1 \mathrm{RT} \\ & \mathrm{P}_2 \mathrm{~V}=\mathrm{n}_2 \mathrm{RT} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{3}{2} \end{aligned} $
$ \begin{aligned} & \mathrm{P}_1=\frac{3}{2} \mathrm{P}_2 ........(ii) \\ & \mathrm{P}_2=\frac{3}{2} \mathrm{P}_2-\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \frac{\mathrm{P}_2}{2}=\frac{\mathrm{kL}}{10 \mathrm{~A}} \\ & \mathrm{P}_2=\frac{\mathrm{kL}}{5 \mathrm{~A}}=\frac{\mathrm{kL}}{\mathrm{~A}} \alpha \\ & \alpha=\frac{1}{5}=0.2 \end{aligned} $
Two identical plates P and Q , radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures $\mathrm{T}_{\mathrm{P}}$ and $\mathrm{T}_{\mathrm{Q}}$, respectively, with $\mathrm{T}_{\mathrm{Q}}<\mathrm{T}_{\mathrm{P}}$, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is $W_0$. Subsequently, two more plates, identical to P and Q , are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from $P$ to $Q$ (Fig. 2) in the steady state is $W_S$, then the ratio $\frac{W_0}{W_S}$ is ________.
Explanation:
Since the plates are blackbodies, they emit radiation according to the Stefan-Boltzmann law. The power emitted per unit area by a blackbody is given by.
$ P=\sigma T^4 . $
Where, $\sigma=$ Stefan - Boltzmann constant
$ \sigma=5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{k}^4 $
When two blackbodies face each other, the net power transfer per unit area from $P$ to $Q=$ power emitted by $P$ toward $Q$ - power emitted by $Q$ toward $P$
So,
$ W_0=\sigma T_p^4-\dot{\sigma} T_a^4=\sigma\left(T_p^4-T_a^4\right)\,\,\,\,\,\,\,\,\,\,\,\,...(i) $
After introducing two additional plates: assuming temperature $T_1$ and $T_2$ for these plates.
In the steady state, the power transfer per unit area through each interface (From $P$ to plate $1,1 \rightarrow 2,2 \rightarrow Q$ ) must be same, because there is no accumulation of energy in the intermediate plates.
So
$ \begin{aligned} & \sigma\left(T_p^4-T_1^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \sigma\left(T_1^4-T_2^4\right)=W_S \,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $
and $\sigma\left(T_2^4-T_a^4\right)=W_S\,\,\,\,...(iv)$
by adding $e q^n$ (2), (3) and (4),
$ \begin{aligned} & \sigma\left(T_p^4-T _1^4+T_1^4-T_2^4+T_2^4-T_Q^4\right)=3 W_S \\ \Rightarrow & \sigma\left(T_p^4-T_Q^4\right)=3 W_S \\ \Rightarrow & W_0=3 W_S \\ & \Rightarrow \frac{W_0}{W_S}=3 \end{aligned} $
Explanation:
Case - 1
$\begin{aligned} & P-P_0=\Delta P=\frac{4 T}{R} \\ & P=\left(P_0+\frac{4 T}{R}\right) \end{aligned}$
Case-2
$\begin{aligned} & \mathrm{P}_1-\frac{8 \mathrm{P}_0}{27}=\Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1} \\ & \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27} \end{aligned}$
Constant temperature process
$\begin{aligned} & \mathrm{PV}=\mathrm{P}_1 \mathrm{~V}_1 \\ & \left(\mathrm{P}_0+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \frac{4}{3} \pi \mathrm{R}^3=\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}+\frac{8 \mathrm{P}_0}{27}\right) \frac{4}{3} \pi \mathrm{R}_1^3 ;\left(\frac{4 \mathrm{~T}}{\mathrm{R}}\right),\left(\frac{4 \mathrm{~T}}{\mathrm{R}_1}\right) \rightarrow \text { (Neglected) } \\ & \mathrm{R}=\frac{2}{3} \mathrm{R}_1 \Rightarrow \mathrm{R}_1=\frac{3}{2} \mathrm{R} \\ & \Delta \mathrm{P}_1=\frac{4 \mathrm{~T}}{\mathrm{R}_1}=\frac{4 \mathrm{~T}}{3 \mathrm{R}} \times 2=\frac{2}{3} \times(144)=96 \mathrm{~Pa} \end{aligned}$
The specific heat capacity of a substance is temperature dependent and is given by the formula $C=k T$, where $k$ is a constant of suitable dimensions in SI units, and $T$ is the absolute temperature. If the heat required to raise the temperature of $1 \mathrm{~kg}$ of the substance from $-73^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$ is $n k$, the value of $n$ is ________.
[Given: $0 \mathrm{~K}=-273{ }^{\circ} \mathrm{C}$.]
Explanation:
To solve this problem, we need to integrate the heat capacity over the given temperature range because the specific heat capacity is temperature dependent. We are given that the specific heat capacity $C$ is defined as $C = kT$, where $k$ is a constant, and $T$ is the absolute temperature.
The heat required to raise the temperature, $Q$, can be found using the following integral:
$ Q = \int_{T_1}^{T_2} C \, dT $
Given $C = kT$, the integral becomes:
$ Q = \int_{T_1}^{T_2} kT \, dT $
We need to convert the given temperatures from Celsius to Kelvin. The temperatures given are:
- Initial temperature: $-73^{\circ} \mathrm{C}$
- Final temperature: $27^{\circ} \mathrm{C}$
In Kelvin, these temperatures are:
- $T_1 = -73^{\circ} \mathrm{C} + 273 = 200 \, \mathrm{K}$
- $T_2 = 27^{\circ} \mathrm{C} + 273 = 300 \, \mathrm{K}$
Now we can evaluate the integral:
$ Q = k \int_{200}^{300} T \, dT $
Integrating, we get:
$ Q = k \left[ \frac{T^2}{2} \right]_{200}^{300} $
Substituting the limits of integration:
$ Q = k \left[ \frac{300^2}{2} - \frac{200^2}{2} \right] $
Solving the values inside the brackets:
$ Q = k \left[ \frac{90000}{2} - \frac{40000}{2} \right] $
$ Q = k \left[ 45000 - 20000 \right] $
$ Q = k \cdot 25000 $
We are given that this heat is equal to $nk$:
$ nk = k \cdot 25000 $
Dividing both sides by $k$, we get:
$ n = 25000 $
Thus, the value of $n$ is 25000.
Explanation:
$\begin{aligned} W_1 & =W_a+W_b+W_c+W_d \\\\ & =4 P_0\left(2 V_0-V_0\right)+n R T \ln \left(\frac{4 V_0}{2 V_0}\right)+2 P_0\left(V_0-4 V_0\right)+0 \\\\ & =4 P_0 V_0+n R\left(\frac{8 P_0 V_0}{n R}\right) \ln 2-6 P_0 V_0 \\\\ & =8 P_0 V_0 \ln 2-2 P_0 V_0\end{aligned}$
$\begin{aligned} W_{\text {II }} & =W_a^{\prime}+W_b^{\prime}+W_c^{\prime}+W_d^{\prime} \\\\ & =n R T \ln \left(\frac{2 V_0}{V_0}\right)+0+P_0\left(V_0-2 V_0\right)+0 \\\\ & =n R\left(\frac{4 P_0 V_0}{n R}\right) \ln 2-P_0 V_0 \\\\ & =4 P_0 V_0 \ln 2-P_0 V_0\end{aligned}$
$\frac{W_I}{W_{I I}}=\frac{8 P_0 V_0 \ln 2-2 P_0 V_0}{4 P_0 V_0 \ln 2-P_0 V_0}=2$
Explanation:
For a gas R and M are constant. So, $\rho T$ = Constant (for constant pressure).
The density of hot air inside the furnace is = $\rho $
The air gets heated inside the furnace at constant pressure Pa.
$ \therefore $ $ \rho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=\rho \mathrm{T} $
$ \Rightarrow $ 1.2 $ \times $ 300 = $\rho $ $ \times $ 360
$\Rightarrow \rho=1 \mathrm{~kg} / \mathrm{m}^3$
Buoyant force applied on the hot air = $\rho_{\mathrm{a}}$Vg (Upward direction)
Weight of the hot air = $\rho $Vg (Downward direction)
$ \therefore $ Net force on the hot air = $\rho_{\mathrm{a}}$Vg - $\rho $Vg
Let acceleration of the hot air in the upward direction = $a$
and mass of the hot air = $\rho $V
$ \therefore $ $\rho $V$a$ = $\rho_{\mathrm{a}}$Vg - $\rho $Vg
$ \Rightarrow $ $a$ = ${{{\rho _a}Vg - \rho Vg} \over {\rho V}}$
= ${{{\rho _a}g - \rho g} \over \rho }$
= ${{1.2 \times 10 - 1 \times 10} \over 1}$
= 2 m/s2
$ \therefore $ Velocity(v) of the hot air when exiting the chimney using formula ${v^2} = {u^2} + 2ah$,
${v^2} = 0 + 2 \times 2 \times 9$
$ \Rightarrow $ v = 6 m/s
Mass flow rate = ${{dm} \over {dt}}$ = $\rho $Av
= $ \rho \times \frac{\pi \mathrm{d}^2}{4} \times\mathrm{v}=1 \times \frac{\pi}{4} \times 10^{-2} \times 6 $
= ${{471} \over {10000}}$ kg/s = ${{471} \over {10000}} \times 1000$ gm/s = 47.1
Explanation:
For a gas R and M are constant. So, $\rho T$ = Constant (for constant pressure).
The density of hot air inside the furnace is = $\rho $
The air gets heated inside the furnace at constant pressure Pa.
$ \therefore $ $ \rho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=\rho \mathrm{T} $
$ \Rightarrow $ 1.2 $ \times $ 300 = $\rho $ $ \times $ 360
$\Rightarrow \rho=1 \mathrm{~kg} / \mathrm{m}^3$
After chimney is closed,
Pressure at the bottom surface, P1 = Pa - $\rho $gh = Pa - (1)(10)(9)
Pressure at the bottom surface, P2 = Pa - $\rho $ag(h + H) = Pa - (1.2)(10)(9 + 1)
$ \therefore $ Pressure difference $\Delta P$ develops between the top and the bottom surfaces of the cap
= P1 - P2
= Pa - (1)(10)(9) - Pa + (1.2)(10)(9 + 1)
= 120 - 90 = 30
Explanation:
For a monatomic ideal gas, the heat capacity at constant volume, $C_{v1}$, is $\left(\frac{3}{2}\right)R$. This comes from the degrees of freedom for a monatomic gas, which are three (x, y, and z motions).
For a diatomic ideal gas, the heat capacity at constant volume, $C_{v2}$, is $\left(\frac{5}{2}\right)R$. This comes from the degrees of freedom for a diatomic gas, which are five (x, y, and z motions and rotation about two axes).
Next, we find the average heat capacity at constant volume for the gas mixture, $C_{v_{\text{mix}}}$, which is a weighted average based on the number of moles of each gas :
$C_{v_{\text{mix}}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2 \times \left(\frac{3}{2}\right)R + 1 \times \left(\frac{5}{2}\right)R}{2 + 1} = \frac{11R}{6}$
The change in internal energy $\Delta U$ for a given number of moles $n$ and change in temperature $\Delta T$ is given by :
$\Delta U = n C_{v} \Delta T$
Given that the work done by the system at constant pressure is :
$W = n R \Delta T$
We can replace $\Delta T = \frac{W}{nR}$ in the equation for $\Delta U$ to get :
$\Delta U = n C_{v_{\text{mix}}} \times \frac{W}{n R} = C_{v_{\text{mix}}} \times \frac{W}{ R} = \frac{11R}{6} \times \frac{66}{ R} = 121 \, \text{Joule}$
So the change in internal energy is indeed 121 Joule.
The value of X is _______________.
Explanation:
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
The value of Y is _______________.
Explanation:
${{Mass} \over {total\,volume}}$ = 1 gm/cc
$ \Rightarrow {{5gm} \over {total\,volume}}$ = 1 gm/cc
$\Rightarrow$ Total volume = 5 cc
$\Rightarrow$ Volume of tube + final volume of air in the tube = 5 cc
$ \Rightarrow {{5gm} \over {2.5gm/cc}} + {V_f} = 5$
$\Rightarrow$ Vf = 5 $-$ 2 = 3 cc
$\Rightarrow$ $\Delta$V = 0.3 cc
For isothermal process,
${p_i}{V_i} = {p_f}{V_f} \Rightarrow {p_f} = {10^5} \times {{3.3} \over 3}$
pf = 1.1 $\times$ 105
pf $-$ pi = 1.1 $\times$ 105 $-$ 105
= 0.1 $\times$ 105 = 10 $\times$ 103 Pa $\Rightarrow$ Y = 10
Explanation:
Ti = 200 K, e = 1
$ - Ms{{dT} \over {dt}} = {{dQ} \over {dt}} = \sigma eA{T^4}$
$ - {{dT} \over {dt}} = {{\sigma A{T^4}} \over {Ms}}$
${{\sigma A} \over {Ms}}\int\limits_{{t_i}}^{{t_f}} {dt = - \int\limits_{{T_i}}^{{T_f}} {{{dT} \over {{T^4}}}} } $
${{\sigma A} \over {Ms}}({t_f} - {t_i}) = {1 \over 3}\left( {{1 \over {T_f^3}} - {1 \over {T_i^3}}} \right)$
${{{{\sigma A} \over {Ms}}({t_1} - 0) = {1 \over 3}\left( {{1 \over {{{(100)}^3}}} - {1 \over {{{(200)}^3}}}} \right)} \over {{{\sigma A} \over {Ms}}({t_2} - 0) = {1 \over 3}\left( {{1 \over {{{(50)}^3}}} - {1 \over {{{(200)}^3}}}} \right)}}$
${{{t_1}} \over {{t_2}}} = {{{{{{(200)}^3} - {{(100)}^3}} \over {{{(100)}^3}{{(200)}^3}}}} \over {{{{{(200)}^3} - {{(50)}^3}} \over {{{(50)}^3}{{(200)}^3}}}}}$
$\therefore$ ${{{t_2}} \over {{t_1}}} = {9 \over 1}$
(take the acceleration due to gravity = 10 ms−2 and the universal gas constant = 8.3 J mol−1K−1).
Explanation:
Volumes of two compartments are
$ V_1=(4+x) A \text { and } V_2=(4-x) A $
At equilibrium,
$ F_2=F_1+m g $
$ \begin{aligned} & P_2 A=P_1 A+m g \\\\ & P_2=P_1+\frac{m g}{A} \end{aligned} $
From $P V=n R T$, we get $P=\frac{n R T}{V}$
$ \begin{aligned} & \frac{n R T}{V_2}=\frac{n R T}{V_1}+\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{V_2}-\frac{1}{V_1}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{1}{A(4-x)}-\frac{1}{4(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & \frac{n R T}{A}\left[\frac{(4+x)-(4-x)}{(4-x)(4+x)}\right]=\frac{m g}{A} \\\\ \Rightarrow & n R T\left[\frac{2 x}{\left(16-x^2\right)}\right]=m g \\\\ \Rightarrow & 0.1 \times 8.3 \times 300\left[\frac{2 x}{\left(16-x^2\right)}\right]=8.3 \times 10 \\\\ \Rightarrow & \frac{6 x}{16-x^2}=1 \Rightarrow 16-x^2=6 x \\\\ \Rightarrow & x^2+6 x-16=0 \\\\ \Rightarrow & x=\frac{-6 \pm \sqrt{36+64}}{2} \\\\ & x=\frac{-6 \pm \sqrt{100}}{2}=\frac{-6 \pm 10}{2} \\\\ & x=\frac{10-6}{2} \text { or } \frac{-10-6}{2}=-8 \text { or } 2 \end{aligned} $
Neglecting negative $\operatorname{sign} x=2$
Partition from top $=4+2=6 \mathrm{~m}$
Explanation:
$ \therefore $ Wexternal + Wwater + Wgas = 0 ....(i)
where Wexternal = work done by external agent
Wwater = work done by water
Wgas = work done by gas
For adiabatic process, ${P_1}V_1^\gamma = {P_2}V_2^\gamma $
Here, ${P_0}{\left[ {{4 \over 3}\pi {R^3}} \right]^{{{41} \over {30}}}} = P{\left[ {{4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]^{{{41} \over {30}}}}$
$ \Rightarrow $ P = ${P_0}{\left[ {{R \over {R - a}}} \right]^{{{41} \over {10}}}}$
Now, Wgas = ${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - P \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3} - {P_0}{{\left[ {{R \over {R - a}}} \right]}^{{{41} \over {10}}}} \times {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \over {{{41} \over {30}} - 1}}$
= ${{{P_0} \times {4 \over 3}\pi {R^3}\left[ {1 - {{\left( {{R \over {R - a}}} \right)}^{{{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]} \over {{{11} \over {30}}}}$
= ${{{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {{{R - a} \over R}} \right)}^{ - {{41} \over {10}}}}{{\left( {{{R - a} \over R}} \right)}^3}} \right]}$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - {{\left( {1 - {a \over R}} \right)}^{ - {{11} \over {10}}}}} \right]$
= ${{40} \over {11}}{P_0}\pi {R^3}\left[ {1 - 1 - {{11a} \over {10R}} - {{\left( { - {{11} \over {10}}} \right)\left( { - {{11} \over {10}} - 1} \right)} \over 2}{{{a^2}} \over {{R^2}}}} \right]$
= -4P0$\pi $R2$a$ - ${{40 \times 21} \over {100 \times 2}}{P_0}\pi R{a^2}$
= -4P0$\pi $R2$a$ - $4.2{P_0}\pi R{a^2}$
Wwater = P0dV
= P0${\left[ {{4 \over 3}\pi {R^3} - {4 \over 3}\pi {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {{R^3} - {{\left( {R - a} \right)}^3}} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ {R - \left( {R - a} \right)} \right]\left[ {{R^2} + R\left( {R - a} \right) + {{\left( {R - a} \right)}^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left[ a \right]\left[ {3{R^2} - 3Ra + {a^2}} \right]} \right]}$
= ${{{4{P_0}} \over 3}\pi \left[ {\left( {3{R^2}a - 3R{a^2} + {a^3}} \right)} \right]}$
= ${4{P_0}\pi \left[ {{R^2}a - R{a^2}} \right]}$ [ignore ${{a^3}}$ term as no ${{a^3}}$ term in the question]
$ \therefore $ Wgas + Wwater = -$4{P_0}\pi R{a^2}$ - $4.2{P_0}\pi R{a^2}$
= $ - 4{P_0}\pi R{a^2}\left[ {1 + 1.05} \right]$
= $ - 4{P_0}\pi R{a^2}\left[ {2.05} \right]$
From (i), Wexternal = - (Wgas + Wwater)
= $4{P_0}\pi R{a^2}\left[ {2.05} \right]$
$ \therefore $ X = 2.05
Explanation:
For small temperature change,
${{dQ} \over {dt}} = e\sigma A{T^3}\Delta T$ .... (i)
${{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T $
$\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T$
${{e\sigma A{T^3}} \over {mC}}$ $\to$ constant for Newton law of cooling
$ \therefore $ ${{e\sigma A{T^3}} \over {mC}} = 0.001 $
$\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001$
$e\sigma A{T^3} = 4.2$ .... (ii)
${{dQ} \over {dt}} = 700 \times 0.05 = 35$ W ..... (iii)
Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get
$35 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33$
Explanation:
${{{p_1}} \over 4}{(4{V_1})^{5/3}} = {p_2}{(32{V_1})^{5/3}}$
${p_2} = {{{p_1}} \over 4}{\left( {{1 \over 8}} \right)^{5/3}} = {{{p_1}} \over {128}}$
${W_{adi}} = {{{p_1}{V_1} - {p_2}{V_2}} \over {\gamma - 1}}$
$ = {{{p_1}{V_1} - {{{p_1}} \over {128}}(32{V_1})} \over {{5 \over 3} - 1}}$
$ = {{{p_1}{V_1}(3/4)} \over {2/3}} = {9 \over 8}{p_1}{V_1}$
${W_{iso}} = {p_1}{V_1}\ln \left( {{{4{V_1}} \over {{V_1}}}} \right) = 2{p_1}{V_1}\ln 2$
$ \therefore $ ${{{W_{iso}}} \over {{W_{adi}}}} = {{16} \over 9}\ln 2$
$ \Rightarrow f = {{16} \over 9} = 1.7778 \approx 1.78$
Explanation:
The degrees of freedom of a monatomic gas is f = 3. Its molar specific heats are ${C_v} = {f \over 2}RT$ and ${C_p} = {C_v} + R = {{(f + 2)} \over 2}RT$. The ratio of specific heats is given by
$\gamma = {C_p}/{C_v} = (f + 2)/f = 5/3$.
In an adiabatic process, $TV_{}^{\gamma - 1}$ = constant. Hence, ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$, which gives
${T_2} = {T_1}{({V_1}/{V_2})^{\gamma - 1}} = 100{(1/8)^{5/3 - 1}} = 25\,K$,
where we used T1 = 100 K and V2 = 8V1. The change in internal energy of one mole of the ideal gas is
$\Delta U = {U_2} - {U_1} = {f \over 2}R{T_2} - {f \over 2}R{T_1} = {f \over 2}R({T_2} - {T_1})$
$ = (3/2)(8)(25 - 100) = - 900\,J$.
Thus, the decrease in internal energy of the gas is 900 J.
Explanation:
Therefore, ${{{p_1}} \over {{p_0}}} = 2$
According to Stefan's law, p $ \propto $ T2
$ \Rightarrow {{{p_2}} \over {{p_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{{2767 + 273} \over {487 + 273}}} \right)^4} = {4^4}$
${{{p_2}} \over {{p_1}}} = {{{p_2}} \over {2{p_0}}} = {4^4} \Rightarrow {{{p_2}} \over {{p_0}}} = 2 \times {4^4}$
${\log _2}{{{p_2}} \over {{p_0}}} = {\log _2}[2 \times {4^4}]$
$ = {\log _2}2 + {\log _2}{4^4}$
$ = 1 + {\log _2}{2^8} = 1 + 8 = 9$
Explanation:
Power, $P = (\sigma {T^4}A) = \sigma {T^4}(4\pi {R^2})$
or, $P \propto {T^4}{R^2}$ ..... (i)
According to Wien's law,
$\lambda \propto {1 \over T}$
($\lambda$ is the wavelength at which peak occurs)
$\therefore$ Eq. (i) will become,
$P \propto {{{R^2}} \over {{\lambda ^4}}}$
or, $\lambda \propto {\left[ {{{{R^2}} \over P}} \right]^{1/4}}$
$ \Rightarrow {{{\lambda _A}} \over {{\lambda _B}}} = {\left[ {{{{R_A}} \over {{R_B}}}} \right]^{1/2}}{\left[ {{{{P_B}} \over {{P_A}}}} \right]^{1/4}}$
$ = {[400]^{1/2}}{\left[ {{1 \over {{{10}^4}}}} \right]^{1/4}} = 2$
Explanation:
The first law of thermodynamics for the process iaf gives
${Q_{iaf}} = {U_{iaf}} + {W_{iaf}} = ({U_f} - {U_i}) + ({W_{ia}} + {W_{af}})$ ..... (1)
Substitute Qiaf = 500 J, Ui = 100 J, Wia = 0 (constant volume), and Waf = 200 J in equation (1) to get Uf = 400 J.
In the process ib,
${Q_{ib}} = {U_{ib}} + {W_{ib}} = ({U_b} - {U_i}) + {W_{ib}}$ ..... (2)
Substitute Ub = 200 J, Ui = 100 J, and Wib = 50 J in equation (2) to get Qib = 150 J.
In the process bf,
${Q_{bf}} = {U_{bf}} + {W_{bf}} = ({U_f} - {U_b}) + {W_{bf}}$ ..... (3)
Substitute Uf = 400 J, Ub = 200 J and Wbf = 100 J in equation (3) to get Qbf = 300 J. Thus, Qbf/Qib = 300/150 = 2.
Explanation:
Change in length $\Delta L = L\,\alpha \,\Delta T$ ..... (i)
Also $Y = {{mgL} \over {A\Delta L}} \Rightarrow \Delta L = {{mgL} \over {YA}}$ ..... (ii)
Equation (i) and (ii) we get ($\because$ $A = \pi {r^2}$)
$m = {{\alpha \Delta TY \times \pi {r^2}} \over g}$
$ = {{({{10}^{ - 5}}) \times (10) \times ({{10}^{11}}) \times 3.14 \times {{(1 \times {{10}^{ - 3}})}^2}} \over {9.8}}$
= 3.2 kg $ \simeq $ 3 kg
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$
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$
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 $0^{\circ} \mathrm{C}$.
One mole of a monatomic ideal gas undergoes the cyclic process $\mathrm{J} \rightarrow \mathrm{K} \rightarrow \mathrm{L} \rightarrow \mathrm{M} \rightarrow \mathrm{J}$, as shown in the P-T diagram.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
[ $\mathcal{R}$ is the gas constant.]
| List-I | List-II |
|---|---|
| (P) Work done in the complete cyclic process | (1) $RT_0 - 4RT_0 \ln 2$ |
| (Q) Change in the internal energy of the gas in the process JK | (2) $0$ |
| (R) Heat given to the gas in the process KL | (3) $3RT_0$ |
| (S) Change in the internal energy of the gas in the process MJ | (4) $-2RT_0 \ln 2$ |
| (5) $-3RT_0 \ln 2$ |
[Given: Wien's constant as $2.9 \times 10^{-3} \mathrm{~m}-\mathrm{K}$ and $\frac{h c}{e}=1.24 \times 10^{-6} \mathrm{~V}-\mathrm{m}$ ]
| List - I | List - II |
|---|---|
| (P) $2000 \mathrm{~K}$ | (1) The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \mathrm{eV}$. |
| (Q) $3000 \mathrm{~K}$ | (2) The radiation at peak wavelength is visible to human eye. |
| (R) $5000 \mathrm{~K}$ | (3) The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction. |
| (S) $10000 \mathrm{~K}$ | (4) The power emitted per unit area is $1 / 16$ of that emitted by a blackbody at temperature $6000 \mathrm{~K}$. |
| (5) The radiation at peak emission wavelength can be used to image human bones. |
List I describes thermodynamic processes in four different systems. List II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.
| List-I | List-II |
|---|---|
| (I) $10^{-3} \mathrm{~kg}$ of water at $100^{\circ} \mathrm{C}$ is converted to steam at the same temperature, at a pressure of $10^{5} \mathrm{~Pa}$. The volume of the system changes from $10^{-6} \mathrm{~m}^{3}$ to $10^{-3} \mathrm{~m}^{3}$ in the process. Latent heat of water $=2250\, \mathrm{~kJ} / \mathrm{kg}$. |
(P) $2 \mathrm{~kJ}$ |
| (II) $0.2$ moles of a rigid diatomic ideal gas with volume $V$ at temperature $500 \mathrm{~K}$ undergoes an isobaric expansion to volume $3 \mathrm{~V}$. Assume $R=8.0 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$. |
(Q) $7 k J$ |
| (III) One mole of a monatomic ideal gas is compressed adiabatically from volume $V=\frac{1}{3} \mathrm{~m}^{3}$ and pressure $2 \mathrm{kPa}$ to volume $\frac{V}{8}$. |
(R) $4 \mathrm{~kJ}$ |
| (IV) Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \mathrm{~kJ}$ of heat and undergoes isobaric expansion. |
(S) $5 \mathrm{~kJ}$ |
| (T) $3 \mathrm{~kJ}$ |
Which one of the following options is correct?
| LIST - I | LIST - II | ||
|---|---|---|---|
| P. | In process I | 1. | Work done by the gas is zero |
| Q. | In process II | 2. | Temperature of the gas remains unchanged |
| R. | In process III | 3. | No heat is exchanged between the gas and its surroundings |
| S. | In process IV | 4. | Work done by the gas is 6P0V0 |