Heat and Thermodynamics

Given, Temperature of food material in refrigerator,

$T_2=4^{\circ} \mathrm{C}=273+4=277 \mathrm{~K}$

Temperature of environment, $T_1=15^{\circ} \mathrm{C}$

$=273+15=288 \mathrm{~K}$

$\therefore$ Carnot efficiency, $\eta=1-\frac{T_2}{T_1}=1-\frac{277}{288}=0.038$

For $\mathrm{N}_2$, degree of freedom $f=5$ [For di-atomic gas]

$\therefore \quad C_{V_1}=\frac{f}{2} \cdot R=\frac{5}{2} R$ and $C_{p_1}=C_{V_1}+R=\frac{7}{2} R$

Amount of $\mathrm{N}_2$ gas, $m_1=7 \mathrm{~g}$

$\therefore$ Number of moles of $\mathrm{N}_2$,

$\begin{aligned} & n_1=\frac{m_1}{28}=\frac{7}{28} \\ & n_1=\frac{1}{4} \end{aligned}$

For Ar, degree of freedom, $f=3$

$\begin{aligned} & \therefore \quad C_{V_2}=\frac{3}{2} R \\ & \therefore \quad C_{p_2}=C_{V_2}+R \\ & =\frac{3}{2} R+R=\frac{5}{2} R \\ & \end{aligned}$

Amount of Ar gas, $m_2=20 \mathrm{~g}$

$\therefore$ Number of moles of Ar gas,

$\begin{aligned} n_2 & =\frac{m_2}{40}=\frac{20}{40}=\frac{1}{2} \\ \therefore \quad\left(C_p\right)_{\text {mix }} & =\frac{n_1 C_{p_1}+n_2 C p_2}{n_1+n_2} \\ & =\frac{\frac{1}{4} \times \frac{7}{2} R+\frac{1}{2} \times \frac{5}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{17 R}{6} \\ \left(C_V\right)_{\text {mix }} & =\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \\ & =\frac{\frac{1}{4} \times \frac{5}{2}+\frac{1}{2} \times \frac{3}{2} R}{\frac{1}{4}+\frac{1}{2}}=\frac{11 R}{6} \\ \therefore \quad\left(\frac{C_p}{C_V}\right)_{\text {mix }} & =\gamma_{\text {mix }}=\frac{\frac{17 R}{6}}{\frac{11R}{6}}=\frac{17}{11} \end{aligned}$

Given, for isobaric process,

Work done by a di-atomic gas

By ideal gas equation, at constant pressure,

$p \Delta V=n R \Delta T$ .... (i)

$\therefore$ Work done,

$\begin{aligned} & W=p \Delta V \\ & W=n R \Delta T \quad \text{.... (ii)} \end{aligned}$

At constant pressure, heat given to gas is given by

$Q=n C_p \Delta T$ ..... (iii)

where, $C_p=$ specific heat at constant pressure

From Eqs. (ii) and (iii), we get

$\frac{W}{Q}=\frac{n R \Delta T}{n C_p \Delta T}=\frac{R}{C_p}$

$\begin{aligned} & =\frac{R}{\frac{(f+2) R}{2}} \quad\left[\because C_p=\frac{(F+2) R}{2}\right] \\ \frac{W}{Q} & =\frac{2}{f+2} \quad[\because W=10 \mathrm{~J} \text { (given) }] \\ \frac{10}{Q} & =\frac{2}{5+2} \quad[\text { For di-atomic gas, } f=5] \\ \Rightarrow \quad Q & =35 \mathrm{~J} \end{aligned}$

Temperature of sphere, $T=227^{\circ} \mathrm{C}$

$=273+227=500 \mathrm{~K}$

Radius, $\quad r=2 \mathrm{~m}$

Emissivity, $e=0.8$

$\therefore$ Radiation power of sphere $=$ Radiated energy per second

$\begin{aligned} & =\sigma A e T^4 \\ & =5.67 \times 10^{-8} \times 4 \pi \times 2^2 \times 0.8 \times(500)^4 \\ & =142502.6 \mathrm{~W}=142.5 \mathrm{~kW} \end{aligned}$

Given, initial pressure of ideal gas,

$p_1=1 \text { bar }=1 \mathrm{~N} / \mathrm{m}^2$

Initital volume $V_1=30 \mathrm{~m}^3$

Final volume, $V_2=10 \mathrm{~m}^3$

Initial temperature, $T_1=320 \mathrm{~K}$

Final temperature, $T_2=280 \mathrm{~K}$

Final pressure of gas, $p_2=$ ?

By ideal gas equation,

$\begin{aligned} & \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2} \Rightarrow \frac{1 \times 30}{320}=\frac{p_2 \times 10}{280} \\ \Rightarrow \quad & p_2=2.625 \mathrm{~bar} \end{aligned}$

There are two vibrational degree of freedom of di-atomic gas molecules at high temperature. Hence, vibrational degree of freedom of a di-atomic gas molecule appears at every high temperature because vibration in gas molecules is directly proportional to the square root of its temperature. Hence, both assertion and reason are true but reason is not the correct explanation of assertion.

A gas can be liquified by applying pressure only when it is cooled below critical temperature. Critical temperature of $\mathrm{NH}_3$ is more than $\mathrm{CO}_2$, i.e., $T_{\mathrm{NH}_3}=405 \mathrm{~K}$ and $T_{\mathrm{CO}_2}=304.1 \mathrm{~K}$.

Therefore, $\mathrm{NH}_3$ is liquified more easily than $\mathrm{CO}_2$. Hence, both assertion and reason are true and Reason is the correct explanation of assertion.

In adiabatic process, heat transfer to the thermodynamic system is zero, i.e., $Q=0$.

Hence, By first law of thermodynamics,

$\begin{array}{ll} & Q=0=\Delta U+W \\ \therefore & W=-\Delta U \end{array}$

So, work done is equal to negative of change in internal energy.

Since, the internal energy is a function of thermodynamic state of the system.

It does not depend on the path, therefore workdone is independent of path in adiabatic process.

Hence, both the assertion and reason are true and reason is the correct explanation of assertion.

Initial and final states are same in all the process.

Hence, $\Delta U=0$ is same for each case.

$\therefore \Delta Q=\Delta W$

Area enclosed by curve with volume.

$\begin{array}{cc} \because & (\text { Area })_1<(\text { Area })_2<(\text { Area })_3 \\\\ \therefore & Q_1< Q_2< Q_3 \end{array}$

For diatomic gas, $\gamma=1.4=\frac{7}{5}$

For adiabatic process, $T V^{\gamma-1}=$ constant

$\begin{aligned} T_1 V_1^{\gamma-1} & =T_2 V_2^{\gamma-1} \\ (300) V_1^{\frac{7}{5}-1} & =T_2\left(\frac{V_1}{5}\right)^{\frac{7}{5}-1} \\ \Rightarrow \quad T_2 & =\frac{300 \times V_1^{2 / 5}}{V_1^{\frac{2}{5}} \times\left(\frac{1}{5}\right)^{2 / 5}}=\frac{300}{5^{\frac{-2}{5}}}=300 \times 5^{2 / 5} \\ & =300 \times 1.903=571 \end{aligned}$

Mean kinetic energy of rotating molecules

$=k T=1.38 \times 10^{-23} \times 571=787.98 \times 10^{-23} \mathrm{~J}$

As there is no change in internal energy of the system during as isothermal change. Hence, the energy taken by the gas is utilised by doing work against external pressure.

Hence, whole heat energy gapping to the body at isothermal process converted into work done. Hence, Assertion is wrong.

Internal energy of gas depends only on temperature of gas.

$\begin{aligned} \text { As, } \alpha_{\text {brass }} & =\frac{\gamma_{\text {brass }}}{3} \\ & =\frac{0.00006}{3}=0.00002 \\ & =2 \times 10^{-5} /{ }^{\circ} \mathrm{C} \end{aligned}$

The brass scale is true at $15^{\circ} \mathrm{C}$, therefore at $30^{\circ}$ its graduations will increase in length and so observed reading will be less than actual reading at $30^{\circ}$.

$\therefore$ The change in reading

$\begin{aligned} \Delta l & =l \alpha_{\text {brass }}(\Delta T)=74.5 \times 2 \times 10^{-5}(30-15) \\ & =0 \cdot 02235 \mathrm{~cm} \end{aligned}$

$\therefore$ Actual reading at $30^{\circ} \mathrm{C}$

$\begin{aligned} l_{30} & =l_{\text {observed }}+\Delta l \\ & =74.5+0.02235 \\ & =74.522 \mathrm{~cm} \end{aligned}$

Assuming area of cross-section to be constant, we have

$V_0 \rho_0=V_{30} \rho_{30} \text { or } a h_0 \rho_0=a h_{30} \rho_{30}$

Therefore, True height at $0^{\circ} \mathrm{C}$

$\begin{aligned} h_0 & =h_{30} \frac{\rho_{30}}{\rho_0}=\frac{h_{30}}{\left(1+\gamma_{\mathrm{Hg}} \Delta T\right)} \\ & =\frac{74.522}{1+0.00018 \times 30} \\ & =\frac{74.522}{1.0054}=74.122 \mathrm{~cm} \end{aligned}$

According to the figure

The change in internal energy of the gas

$\begin{aligned} \Delta U & =\frac{p_2 V_2-p_1 V_1}{\gamma-1} \\ p_1 & =2 p a \\ p_2 & =5 p a \\ V_1 & =4 \mathrm{~m}^3 \\ V_2 & =6 \mathrm{~m}^3 \\ & =\frac{2 \times 6-4 \times 5}{\frac{7}{5}-1} \\ & =-20 \mathrm{~kJ} \end{aligned}$

Over top of a fire heat is transferred by both convection and radiation while on the sides heat is transferred by radiation only.

In adiabatic expansion $\Delta Q=0$

$W=+ \text { ve, } \Delta U=- \text { ve }$

$\therefore U, T$ or product of $p V$ will decrease because $U \propto T \propto p V$

A monomatomic gas molecules like $\mathrm{He}$ consists of single atom. It can have translational motion in any direction in space. Thus, it has three translational degree of freedom $f=3$ (all translational)

It can also rotate but due to its small moment of inertia rotational kinetic energy is neglected. The molecules of a diatomic gas (like $\mathrm{O}_2, \mathrm{CO}_2, \mathrm{H}_2$) cannot only move body but also rotate about any one of the three co-ordinates. Hence it can have two rotational degree of freedom. Thus a diatomic molecule has 5 degree of freedom i.e. 3 translational and 2 rotational.

$C=\frac{Q}{m \Delta T}$

In isothermal process $\Delta T=0$. Therefore, $C$ is not defined further in isothermal process $p V$ and $T$ both are constants.