Heat and Thermodynamics

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

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.