Explanation:
${1 \over 4} = 1 - {{{T_1}} \over {{T_2}}}$
${{{T_1}} \over {{T_2}}} = {3 \over 4}$
When the temperature of the sink is reduced by 52K then its efficiency is doubled.
${1 \over 2} = 1 - {{\left( {{T_1} - 52} \right)} \over {{T_2}}}$
${{{T_1} - 52} \over {{T_2}}} = {1 \over 2}$
${{{T_1}} \over {{T_2}}}{{ - 52} \over {{T_2}}} = {1 \over 2}$
${3 \over 4} - {{52} \over {{T_2}}} = {1 \over 2}$
${{52} \over {{T_2}}} = {1 \over 4}$
${T_2} = 208$ K
Explanation:
To find the change in temperature of a monoatomic gas when the container it is in is suddenly stopped, we can follow these steps:
Relationship Between Kinetic Energy and Internal Energy Change:
The kinetic energy lost by the container when it stops is converted into the internal energy of the gas:
$ \Delta KE = \Delta U $
Expression for Internal Energy Change:
The change in internal energy can be expressed in terms of the change in temperature:
$ \Delta U = n C_v \Delta T $
where $ n $ is the number of moles, $ C_v $ is the molar heat capacity at constant volume, and $ \Delta T $ is the change in temperature.
Using the Kinetic Energy Formula for the Change:
For a monoatomic gas, the molar heat capacity at constant volume $ C_v $ is $\frac{3R}{2}$. Thus, we equate the kinetic energy to the change in internal energy:
$ \frac{1}{2} m v^2 = \frac{3}{2} n R \Delta T $
Solving for $\Delta T$:
Rearrange the equation to solve for the change in temperature:
$ \Delta T = \frac{m v^2}{3 n R} $
Substituting Given Values:
Here, mass $ m = 4 $ u, velocity $ v = 30 $ m/s, and the number of moles $ n = 1 $. Substitute these into the equation:
$ \Delta T = \frac{4 \times (30)^2}{3 \times 1 \times R} $
Simplifying:
$ \Delta T = \frac{3600}{R} $
Relating to Given Expression:
According to the problem, $\Delta T = \frac{x}{3R}$. Therefore, equating:
$ \frac{x}{3R} = \frac{3600}{R} $
Solve for $x$:
Simplifying gives:
$ x = 3600 $
Thus, the value of $x$ is 3600.
Explanation:
In this thermodynamical process, the pressure $ P $ of a gas is related to its volume $ V $ by the equation $ P = kV^3 $. Thus, we have:
$ PV^{-3} = k $
This implies that the value of $ x $ is $-3$ in the relationship. To find the work done ($ W $) when the temperature changes from 100°C to 300°C, we can use the formula:
$ W = \frac{nR(T_1 - T_2)}{x - 1} $
Substitute the given temperatures and the value of $ x $:
$ W = \frac{nR(100 - 300)}{-3 - 1} $
Calculate the expression:
$ W = \frac{nR(-200)}{-4} $
Which simplifies to:
$ W = 50 \, nR $
Explanation:
As we know that,
Root mean square speed, ${v_{rms}} = \sqrt {{{3RT} \over m}} $
$\therefore$ ${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $
$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$ ms$-$1
$ \Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$
Explanation:
Work = Area of ABCD = (2P0)(V0)
Qin = QAB + QBC
QAB = isochoric process
= nCV(TB - TA)
= 1 $ \times $ ${3 \over 2}R\left( {{T_B} - {T_A}} \right)$
= ${3 \over 2}R\left( {{{3{P_0}{V_0}} \over R} - {{{P_0}{V_0}} \over R}} \right)$
= 3P0V0
QBC = isobaric process
= nCP$\Delta $T
= $1 \times {5 \over 2}R$(TC - TB)
= ${5 \over 2}R\left( {{{6{P_0}{V_0}} \over R} - {{3{P_0}{V_0}} \over R}} \right)$
= 7.5 P0V0
$ \therefore $ $\eta $ = ${{{2{P_0}{V_0}} \over {3{P_0}{V_0} + 7.5{P_0}{V_0}}}}$ $ \times $ 100
$ \simeq $ 19%
Explanation:
$ \therefore $ P1V1 = nR (250)
and P2(2V1) = ${{5n} \over 4}R \times \left( {2000} \right)$
By Dividing
${{{P_1}} \over {2{P_2}}}$ = ${{4 \times 250} \over {5 \times 2000}}$
$ \Rightarrow $ ${{{P_1}} \over {{P_2}}} = {1 \over 5}$
$ \Rightarrow $ ${{{P_2}} \over {{P_1}}} = 5$
(Molar mass of N2 gas 28 g).
Explanation:
VN2 = $\sqrt {{{3R(573)} \over 28}} $
VH2 = $\sqrt {{{3RT} \over 2}} $
Given, VN2 = VH2
$\sqrt {{{3RT} \over 2}} $ = $\sqrt {{{3R(573)} \over 28}} $
$ \Rightarrow $ ${T \over 2} = {{573} \over {28}}$
$ \Rightarrow $ T = 41 K
If |$\Delta $T| = C|$\Delta $P| then value of C in (K/atm.) is _________.
Explanation:
$ \therefore $ $P\Delta V + V\Delta P = 0$ (for constant temp.)
and $P\Delta V$ = $nR\Delta T$ (for constant pressure)
$\Delta T = {{P\Delta V} \over {nR}}$
$\Delta P = - {{P\Delta V} \over V}$ ($\Delta V$ is same in both cases)
${{\Delta T} \over {\Delta P}} = {{P\Delta V} \over {nR}}{V \over { - P\Delta V}} = {{ - V} \over {nR}} = - {T \over P}$
[As PV = nRT
$ \Rightarrow $ $ {{V \over {nR}} = {T \over P}} $]
$ \therefore $ $\left| {{{\Delta T} \over {\Delta P}}} \right| = \left| {{{ - 300} \over 2}} \right| = 150$
Explanation:
$ \therefore $ Total internal energy of gases remain same.
u1 + u2 = u1' + u2'
We know, $\Delta $U = nCv$\Delta $T
$ \therefore $ $\left( {0.1 \times {{3R} \over 2} \times 200} \right) + \left( {0.05 \times {{3R} \over 2} \times 400} \right)$ = $\left( {0.15 \times {{3R} \over 2} \times {T_f}} \right)$
$ \Rightarrow $ (20 + 20) = 0.15 Tf
$ \Rightarrow $ Tf = 266.67
Explanation:
Q2 = W + Q1
Coefficient of performance (C.O.P)
$ = {{{Q_1}} \over W} = {{{Q_1}} \over {{Q_2} - {Q_1}}} = {{{T_1}} \over {{T_2} - {T_1}}}$
${{{Q_1}} \over W} = {{273} \over {300 - 273}}$
${{{Q_1}} \over W} = {{273} \over {27}}$
$W = {{27} \over {273}}{Q_1}$
$W = {{27} \over {273}}mL$
$W = {{27} \over {273}} \times 80 \times 100$
${Q_2} = {{27} \over {273}} \times 80 \times 100 + 80 \times 100$
$ = 8791.2$ cal
Explanation:
$ \therefore $ V1$\gamma $1 = V2$\gamma $2
$ \Rightarrow $ 500 $ \times $ 6 $ \times $ 10-6 = Vm $ \times $ 1.5 $ \times $ 10-4
$ \Rightarrow $ Vm = ${{500 \times 6 \times {{10}^{ - 6}}} \over {1.5 \times {{10}^{ - 4}}}}$
$ \Rightarrow $ Vm = 20 cc
Explanation:
f = 5
$\gamma $ = ${7 \over 5}$
Ti = T = 273 + 20 = 293 K
Vi = V
Vf = ${V \over {10}}$
For adiabatic process TV$\gamma $ - 1 = constant
${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$
$ \Rightarrow $ $\left( {293} \right){V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {10}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ ${T_2} = 293 \times {\left( {10} \right)^{{2 \over 5}}}$
$\Delta $U = ${{nfR\left( {{T_2} - {T_1}} \right)} \over 2}$
= ${{5 \times 5 \times {{25} \over 3} \times \left( {{{293.10}^{{2 \over 5}}} - 293} \right)} \over 2}$
= ${{625 \times 293 \times \left( {{{10}^{{2 \over 5}}} - 1} \right)} \over 6}$
= 46.14 $ \times $ 103 J
$ \simeq $ 46 kJ
ideal diatomic gas ($\gamma $ = 1.4) is first compressed
adiabatically from volume V1 to V2 = ${{{V_1}} \over {16}}$. It is
then allowed to expand isobarically to volume 2V2. If all the processes are the quasi-static then
the final temperature of the gas (in oK) is (to the nearest integer) _____.
Explanation:
$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$
$ \Rightarrow $ T2 = 300 × (16)0.4
Isobaric process
V = ${{nRT} \over P}$
V2 = kT2... (1)
2V 2 = KTf... (2)
Tf = 2T2 = 300 × 2 × (16)0.4 = 1818 K
The value of $\theta $ (in °C to the nearest integer) is ..........
Explanation:
Applying law of calorimetry
1(T1 – 60) + 2(T2 – 60) = 0
$ \Rightarrow $ T1 + 2T2 = 180 ....(1)
1(T2 – 30) + 2(T3 – 30) = 0
$ \Rightarrow $ T2 + 2T3 = 90 ......(2)
2(T1 – 60) + 1(T3 – 60) = 0
$ \Rightarrow $ 2T1 + T3 = 180 .....(3)
from (1), (2), (3)
T1 = 80o C
T2 = 50o C
T3 = 20o C
1 × (T1 – $\theta $) + 1 ×(T2 – $\theta $) + 1(T3 – $\theta $) = 0
$ \Rightarrow $ T1 + T2 + T3 = 3$\theta $
$ \Rightarrow $ $\theta $ = ${{80 + 50 + 20} \over 3}$ = 50oC
Explanation:
$ \Rightarrow $ 600 M = 24000
$ \Rightarrow $ M = 40
5 $ \times $ 10-5/oC along the x-axis and 5 $ \times $ 10-6/oC along the y and the z-axis. If the coefficient of volume expansion of the solid is C $ \times $ 10-6/oC then the value of C is
Explanation:
$ \Rightarrow $ C $ \times $ 10–6 = 5 × 10–5 + 5 × 10–6 + 5 × 10–6
$ \Rightarrow $ C $ \times $ 10–6 = 50 × 10–6 + 10 × 10–6
$ \Rightarrow $ C = 60
Explanation:
So ${{Q + 1200} \over Q} = {{900} \over {300}}$
$ \Rightarrow $ Q + 1200 = 3Q
$ \Rightarrow $ Q = 600 J
Choose the correct statement for processes A & B shown in figure.
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$