Heat and Thermodynamics

From gas equation,

$pV = nRT \Rightarrow p = {{nRT} \over V}$

Pressure of one mole of O₂ gas,

${(p)_{{O_2}}} = {{1 \times RT} \over V}$ ..... (i)

Pressure of one mole of He gas,

${(p)_{He}} = {{1 \times R(2T)} \over V}$ ...... (ii)

Dividing Eq. (ii) by Eq. (i), we get

From Ideal gas equation,

$pV = nnT = {{mRT} \over M}$

$n = {{Molar\,mass\,(m)} \over {Molecular\,weight\,(M)}}$

Here, n = number of moles, R = Gas constant, V = volume

$ \Rightarrow {{pV} \over m} = {{RT} \over M}$

$ \Rightarrow {p \over {(m/V)}} = {{RT} \over M}$

${p \over d} = {{RT} \over m}$ [$\because$ ${m \over v} = d$]

$ \Rightarrow {p \over d} \propto T$

$\Rightarrow$ Slope of p versus d graph $\propto$ T

Since, slope is less for the gas at temperature T₁.

Therefore, T₁ < T₂

Let, $\gamma$ be the adiabatic exponent for the mixture.

For monoatomic gas,

n₁ = 2 and $\gamma$₁ = ${{5} \over {3}}$

For diatomic gas,

n₂ = 3 and $\gamma$₂ = ${{7} \over {5}}$

For the mixture,

${{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}$

Substituting all the given values in above equation, we get

${{2 + 3} \over {\gamma - 1}} = {2 \over {{5 \over 3} - 1}} + {3 \over {{7 \over 5} - 1}}$

$ \Rightarrow {5 \over {\gamma - 1}} = {{21} \over 2} \Rightarrow \gamma = {{31} \over {21}}$

An HCl molecule, being diatomic, has:

• 3 translational degrees of freedom
• 2 rotational degrees of freedom
• 2 vibrational degrees of freedom

The total number of degrees of freedom is $3 + 2 + 2 = 7$.

According to the equipartition theorem, each degree of freedom contributes $\frac{1}{2} k_B T$ to the total energy. So the total energy is given by: $\frac{7}{2} k_B T$

The translational kinetic energy is related to the root-mean-square (rms) speed $\overline{v}$ by: $\frac{1}{2} m \overline{v}^2 = \frac{7}{2} k_B T$

Rearranging to solve for the temperature, we find: $T = \frac{m \overline{v}^2}{7k_B}$

We know, m = V . $\rho$

where, V = volume and $\rho$ = density.

So, we have

${{dQ} \over {dt}} = {h \over {ms}}(T - {T_0}) = {{h(T - {T_0})} \over {V\,.\,\rho s}}$

Since, h, (T $-$ T₀) and V are constant for both beaker.

$\therefore$ ${{dQ} \over {dt}} \propto {1 \over {\rho s}}$

We have given that $\rho$_A = 8 $\times$ 10² kgm^$-$3, $\rho$_B = 10³ kgm^$-$3, s_A = 2000 J kg^$-$1 K^$-$1 and s_B = 4000 J kg^$-$1 K^$-$1, $\rho$_As_A = 16 $\times$ 10⁵ and $\rho$_Bs_B = 4 $\times$ 10⁶

$\rho$_A < $\rho$_B, s_A < s_B and $\rho$_As_A < $\rho$_Bs_B

$ \Rightarrow {1 \over {{\rho _A}{s_A}}} > {1 \over {{\rho _B}{s_B}}} \Rightarrow {{d{Q_A}} \over {dt}} > {{d{Q_B}} \over {dt}}$

So, for container B, rate of cooling is smaller than the container A. Hence, graph of B lies above the graph of A and it is not a straight line (slope of A is greater than B).

Let x grams of water is evaporated.

According to the principle of calorimetry,

Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water

Given, mass of water = 150 g

$ \Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \times 2.10 \times {10^6}$

$ \Rightarrow (150 - x) \times 3.36 = 21x$

$ \Rightarrow x = {{150} \over {7.25}} = 20.6$

$\therefore$ $x \approx 20g$