Heat And Thermodynamics
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General Learning Resources
Total Translational K.E. Of Gas
$=\frac{1}{2} M\left\langle v^{2}\right\rangle=\frac{3}{2} P V=\frac{3}{2} n R T$
$<v^{2}>=\frac{3 P}{\rho} \quad v_{rms}=\sqrt{\frac{3 P}{\rho}}=\sqrt{\frac{3 RT}{M_{mol}}}=\sqrt{\frac{3 KT}{m}}$
Important Points :
$-\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}$
$\bar{v}=\sqrt{\frac{8 K T}{\pi m}}=1.59 \sqrt{\frac{K T}{m}}$
$\mathrm{v}_{\mathrm{rms}}=1.73 \sqrt{\frac{\mathrm{KT}}{\mathrm{m}}}$
Most probable speed:
$V_{p}=\sqrt{\frac{2 KT}{m}}=1.41 \sqrt{\frac{KT}{m}} \therefore V_{rms}>\overline{V}>V_{mp}$
Degree of freedom :
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Monoatomic $f=3$
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Diatomic $f=5$
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Polyatomic $f=6$
Maxwell’s Law Of Equipartition Of Energy :
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Total K.E. of the molecule $=\frac{1}{2} \mathrm{fKT}$
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For an ideal gas : Internal energy $U=\frac{f}{2} n R T$
Ideal gas law:
$PV = nRT$
First Law Of Thermodynamics:
$\Delta U = Q - W$
Isothermal Process:
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Workdone in isothermal process : $W=\left[2.303 nRT \log _{10} \frac{V_f}{V_i}\right]$
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Internal energy in isothermal process : $ \Delta \mathrm{U}=0$
Isochoric Process
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Work done in isochoric process : $ d W=0$
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Change in internal energy for an ideal gas during an isochoric process: $\Delta U = nC_v\Delta T$
Isobaric process :
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Work done: $\Delta \mathrm{W}=n R\left(T_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}\right)$
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Change in internal energy for isobaric process: $\Delta U = n C_p \Delta T$
Adiabatic process :
- Work done: $\Delta W=\frac{nR\left(T_{i}-T_{f}\right)}{\gamma-1}$
Specific heat :
$ C_{V}=\frac{f}{2} R \quad C p=\left(\frac{f}{2}+1\right) R $
Molar heat capacity of ideal gas in terms of $\mathbf{R}$ :
(i) for monoatomic gas : $\frac{C_{p}}{C_{v}}=1.67$
(ii) for diatomic gas : $\frac{C_{p}}{C_{v}}=1.4$
(iii) for triatomic gas : $\frac{C_{p}}{C_{v}}=1.33$
Heat Capacity Ratio :
$\gamma=\frac{C_{p}}{C_{v}}=\left[1+\frac{2}{f}\right]$
Mayer’s equation:
For ideal gas: $C_{p}-C_{v}=R $
In cyclic process :
$\Delta Q=\Delta W$
In a mixture of non-reacting gases :
$\text{Molar weight}=\frac{n_{1} M_{1}+n_{2} M_{2}}{n_{1}+n_{2}}$
$C_{v}=\frac{n_{1} C_{v_{1}}+n_{2} C_{v_{2}}}{n_{1}+n_{2}}$
$\gamma=\frac{C_{p(\text { mix })}}{C_{v(\text { mix })}}=\frac{n_{1} C_{p_{1}}+n_{2} C_{p_{2}}+\ldots}{n_{1} C_{v_{1}}+n_{2} C_{v_{2}}+\ldots}$
Heat Engines
$\text{Efficiency,} \eta=\frac{\text { work done by the engine }}{\text { heat sup plied to it }}$
$\eta=\frac{W}{Q_{H}}=\frac{Q_{H}-Q_{L}}{Q_{H}}=1-\frac{Q_{L}}{Q_{H}}$
Second law of Thermodynamics
- Kelvin- Planck Statement
It is impossible to construct an engine, operating in a cycle, which will produce no effect other than extracting heat from a reservoir and performing an equivalent amount of work.
- Clausius Statement
It is impossible to make heat flow from a body at a lower temperature to a body at a higher temperature without doing external work on the working substance
Entropy:
- Change in entropy of the system is $\Delta S=\frac{\Delta Q}{T} \Rightarrow S_{f}-S_{i}=\int_{i}^{f} \frac{\Delta Q}{T}$
- In an adiabatic reversible process, entropy of the system remains constant.
Efficiency of Carnot Engine:
(1) Operation I (Isothermal Expansion)
(2) Operation II (Adiabatic Expansion)
(3) Operation III (Isothermal Compression)
(4) Operation IV (Adiabatic Compression)
Thermal Efficiency of a Carnot Engine:
$\frac{V_{2}}{V_{1}}=\frac{V_{3}}{V_{4}} \Rightarrow \frac{Q_{2}}{Q_{1}}=\frac{T_{2}}{T_{1}} \Rightarrow \eta=1-\frac{T_{2}}{T_{1}}$
Refrigerator (Heat Pump)
- Coefficient of performance, $\beta=\frac{Q_{2}}{W}=\frac{1}{\frac{T_{1}}{T_{2}}-1}==\frac{1}{\frac{T_{1}}{T_{2}}-1}$
Calorimetry And Thermal Expansion Types Of Thermometers :
(a) Liquid Thermometer : $ T=\left[\frac{\ell-\ell_{0}}{\ell_{100}-\ell_{0}}\right] \times 100$
(b) Gas Thermometer :
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Constant volume : $T=\left[\frac{P-P_{0}}{P_{100}-P_{0}}\right] \times 100 ; P=P_{0}+\rho g h$
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Constant Pressure : $T=\left[\frac{\mathrm{V}}{\mathrm{V}-\mathrm{V}^{\prime}}\right] \mathrm{T}_{0}$
(c) Electrical Resistance Thermometer :
$ T=\left[\frac{R_{t}-R_{0}}{R_{100}-R_{0}}\right] \times 100 $
Thermal Expansion :
(a) Linear :
$ \alpha=\frac{\Delta L}{L_{0} \Delta T} \quad \text { or } \quad L=L_{0}(1+\alpha \Delta T) $
(b) Area/superficial :
$ \beta=\frac{\Delta A}{A_{0} \Delta T} \quad \text { or } \quad A=A_{0}(1+\beta \Delta T) $
(c) volume/ cubical :
$ r=\frac{\Delta V}{V_{0} \Delta T} \quad \text { or } \quad V=V_{0}(1+\gamma \Delta T) $
$ \alpha=\frac{\beta}{2}=\frac{\gamma}{3} $
Thermal stress of a material :
$ \frac{F}{A}=Y \frac{\Delta \ell}{\ell} $
Energy stored per unit volume :
$ E=\frac{1}{2} K(\Delta L)^{2} \quad \text { or } \quad E=\frac{1}{2} \frac{A Y}{L}(\Delta L)^{2} $
Variation of time period of pendulum clocks :
$ \Delta \mathrm{T}=\frac{1}{2} \alpha \Delta \theta \mathrm{T} $
$ \mathrm{T}^{\prime}<\mathrm{T} \quad \text { - clock-fast : time-gain } $
$ \mathrm{T}^{\prime}>\mathrm{T} \quad \text { - clock slow : time-loss } $
Calorimetry :
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Specific heat: $S=\frac{\mathrm{Q}}{\mathrm{m} \cdot \Delta \mathrm{T}}$
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Molar specific heat: $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
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Water equivalent: $m_{m} S_{m}=m_{w} S_{w}$
Heat Transfer
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Thermal Conduction : $ \frac{d Q}{d t}=-K A \frac{d T}{d x}$
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Thermal Resistance : $\mathrm{R}=\frac{\ell}{\mathrm{KA}}$
Series And Parallel Combination Of Rod :
(i) Series : $\frac{\ell_{\text {eq }}}{K_{\text {eq }}}=\frac{\ell_{1}}{K_{1}}+\frac{\ell_{2}}{K_{2}}+\ldots$ when $\left(A_{1}=A_{2}=A_{3}=\ldots\right)$
(ii) Parallel : $K_{\text {eq }} A_{e q}=K_{1} A_{1}+K_{2} A_{2}+\ldots$ when $\left(\ell_{1}=\ell_{2}=\ell_{3}=\ldots\right)$
For absorption, reflection and transmission: $ r+t+a=1 $
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Emissive power : $ \mathrm{E}=\frac{\Delta \mathrm{U}}{\Delta \mathrm{A} \Delta \mathrm{t}}$
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Spectral emissive power : $ E_{\lambda}=\frac{d E}{d \lambda}$
Emissivity : $ e=\frac{E \text { of a body at } \mathrm{T} \text { temperature }}{\mathrm{E} \text { of a black body at } \mathrm{T} \text {temperature}}$
- Kirchoff’s law : $4\frac{E \text { (body) }}{a \text { (body) }}=E \text{(black body)}$
Wein’s Displacement law :
$\lambda_{\mathrm{m}} \cdot \mathrm{T}=\mathrm{b}$
$b=0.282 \mathrm{~cm}-\mathrm{k}$
Stefan Boltzmann law :
$ \mathrm{u}=\sigma \mathrm{T}^{4} \quad \quad \mathrm{~s}=5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{2} \mathrm{k}^{4} $
$ \Delta u=u-u_{0}=e \sigma A \left(T^{4}-T_{0}^{4}\right) $
Newton’s law of cooling :
$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{k}\left(\theta-\theta_{0}\right) ; \quad \theta=\theta_{0}+\left(\theta_{\mathrm{i}}-\theta_{0}\right) \mathrm{e}^{-\mathrm{kt}}$