Heat and Thermodynamics

According to kinetic theory of gases

$ p V=\frac{1}{3} M V_{\mathrm{rms}}^2 $

at constant temperature $V_{\text {rms }}^2$ is constant, as $V$ is also constant, so $p \propto M$

When the mass of gas increases by $50 \%$ pressure also increases by 50\%.

So, final pressure $=76+\frac{50}{100} \times 76=114 \mathrm{~cm}$ of Hg .

$ \begin{aligned} &\text { Since, two cylinders are in parallel combination }\\ &\begin{array}{ll} \therefore & \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2} \\ \Rightarrow & \frac{k_e A_e}{l}=\frac{k_1 A_1}{l}+\frac{k_2 A_2}{l} \\ \Rightarrow & k_e \pi(2 R)^2=k_1 \pi(R)^2+k_2 \pi\left[(2 R)^2-(R)^2\right] \\ \Rightarrow & 4 k_e=k_1+3 k_2 \\ \therefore & k_e=\frac{k_1+3 k_2}{4} \end{array} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Here, \,\, d Q & =400 \mathrm{cal}, d W=-105 \mathrm{~J} \\ & =-\frac{105}{4.2} \mathrm{cal}=-25 \mathrm{cal} ; d U=? \end{aligned} \end{aligned} $

Now, $d U=d Q-d W$

$ d U=400-(-25)=425 \mathrm{cal} $

Note $d W$ is negative because work is done on the system.

$ \begin{aligned} & \text { As, }\left(\frac{\Delta Q}{\Delta t}\right)_P=\left(\frac{\Delta Q}{\Delta t}\right)_Q \\ & \Rightarrow K_1 A_1 \frac{\left(T_1-T_2\right)}{l}=K_2 A_2 \frac{\left(T_1-T_2\right)}{l} \\ & \text { or } \quad K_1 A_1=K_2 A_2 \text { or } \frac{A_1}{A_2}=\frac{K_2}{K_1} \end{aligned} $

In the cycle $ A \rightarrow B \rightarrow C \rightarrow D \rightarrow A $, heat is added to the system during the processes $ D \rightarrow A $ (at constant volume) and $ A \rightarrow B $ (at constant pressure).

Calculating the Heat Added

Heat for Process $ D \rightarrow A $ (Constant Volume):

For an ideal monoatomic gas, the molar heat capacity at constant volume $ C_V $ is given by:

$ C_V = \frac{3}{2} R $

The heat added during this process is:

$ Q_{DA} = n C_V (\Delta T)_{DA} = \frac{3}{2} n R \Delta T $

Heat for Process $ A \rightarrow B $ (Constant Pressure):

The molar heat capacity at constant pressure $ C_p $ is given by:

$ C_p = \frac{5}{2} R $

The heat added during this process is:

$ Q_{AB} = n C_p (\Delta T)_{AB} = \frac{5}{2} n R \Delta T $

Total Heat Supplied

The total heat supplied in the cycle is the sum of the heat added during these two processes:

$ Q = Q_{DA} + Q_{AB} $

Substitute the expressions for $ Q_{DA} $ and $ Q_{AB} $:

$ Q = \frac{3}{2} n R \Delta T + \frac{5}{2} n R \Delta T $

Given $ n R \Delta T = p \Delta V $, and substituting it:

$ Q = \frac{3}{2}(p_0 V_0) + \frac{5}{2}(p_0 V_0) $

This simplifies to:

$ Q = \frac{13}{2} p_0 V_0 $

In a mixture of gases where the average number of degrees of freedom per molecule is 6, the velocity of sound in the gas can be determined using the following relationships:

The root mean square (rms) speed $ v_{\mathrm{rms}} $ is given by:

$ v_{\mathrm{rms}} = \sqrt{\frac{3 R T}{M}} $

The velocity of sound $ v_{\text{sound}} $ is given by:

$ v_{\text{sound}} = \sqrt{\frac{\gamma R T}{M}} $

The ratio of the velocity of sound to the rms speed is:

$ \frac{v_{\text{sound}}}{v_{\mathrm{rms}}} = \sqrt{\frac{\gamma}{3}} $

Given that the degrees of freedom $ f = 6 $, we calculate the adiabatic index $ \gamma $ as follows:

$ \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = \frac{4}{3} $

Substituting $ \gamma = \frac{4}{3} $ into the equation for the velocity of sound:

$ v_{\text{sound}} = \sqrt{\frac{4/3}{3}} \, v_{\mathrm{rms}} = \frac{2}{3} v_{\mathrm{rms}} $

Therefore, the velocity of sound in the gas is:

$ \frac{2c}{3} $

To determine the pressure of the helium (He) gas in the identical vessel, we can use the ideal gas law:

$PV = nRT$

Given that both vessels are identical, they have the same volume, $V$. Also, each vessel contains one mole of gas, so $n = 1$ for both cases. Let's first express the pressure $p$ of the oxygen (O₂) gas in terms of the given conditions:

For the oxygen gas:

$pV = nRT$

Substituting the given values:

$pV = (1 \text{ mole}) \cdot R \cdot T$

Therefore,

$pV = RT$

Now, consider the helium gas in the other vessel at temperature $4T$. According to the ideal gas law:

$P_{He}V = nRT_{He}$

Here, ${T_{He} = 4T}$. Substituting the known values:

$P_{He}V = (1 \text{ mole}) \cdot R \cdot (4T)$

Therefore,

$P_{He}V = 4RT$

Since $pV = RT$ from our earlier equation for the oxygen gas, we can substitute $RT$ with $pV$ in the helium pressure equation:

$P_{He}V = 4(pV)$

Dividing both sides by $V$ to isolate $P_{He}$:

$P_{He} = 4p$

Therefore, the pressure of the helium gas at temperature $4T$ is $4p$.

Thus, the correct answer is:

Option C: 4p

A Carnot heat engine operates between two temperatures: the hot reservoir temperature ($ T_h $) and the cold reservoir temperature ($ T_c $). To find the efficiency ($ \eta $) of a Carnot engine, we use the formula:

$ \eta = 1 - \frac{T_c}{T_h} $

where both temperatures ($ T_h $ and $ T_c $) must be in absolute temperature, i.e., Kelvin (K).

The given temperatures are $ 527^{\circ} \mathrm{C} $ and $ 127^{\circ} \mathrm{C} $. First, we convert these temperatures to Kelvin:

$ T_h = 527^{\circ} \mathrm{C} + 273 = 800 \, \mathrm{K} $

$ T_c = 127^{\circ} \mathrm{C} + 273 = 400 \, \mathrm{K} $

Next, we calculate the efficiency of the Carnot engine:

$ \eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5 $

The efficiency ($ \eta $) is equal to the fraction of heat converted to work. Therefore, the fraction of heat used to do mechanical work per second is:

$ \eta = \frac{W}{Q_h} $

where $ W $ is the work done per second (which is $ 1.0 \, \text{kW} $ or $ 1000 \, \text{W} $), and $ Q_h $ is the heat consumed per second.

Rearranging the formula, we get:

$ Q_h = \frac{W}{\eta} $

Substituting the given values:

$ Q_h = \frac{1000 \, \mathrm{W}}{0.5} = 2000 \, \mathrm{W} $

To convert the heat consumed per second from Watts to kcal/s, we use the conversion factor $ 1 \, \mathrm{kcal/s} = 4184 \, \mathrm{W} $:

$ Q_h = \frac{2000 \, \mathrm{W}}{4184 \, \mathrm{W/kcal}} \approx 0.477 \, \mathrm{kcal/s} $

Therefore, the amount of heat the Carnot engine should consume per second to deliver mechanical work at the rate of $ 1.0 \, \text{kW} $ is:

Option A: 0.477 kcal/s

Let the insulated container has n moles of the monoatomic gas. The loss of kinetic energy of the gas,

$\Delta {E_K} = {1 \over 2}(mn){v^2}$

If the change in temperature of the gas is $\Delta$T, then heat gained by the gas is,

$\Delta Q = {3 \over 2}nR\,\Delta T$

Now, according to question,

$\Delta Q = \Delta {E_K}$

${3 \over 2}nR\,\Delta T = {1 \over 2}mn{v^2}$

$ \Rightarrow \Delta T = {{m{v^2}} \over {3R}}$

In the curve, V $\propto$ $\frac{1}{T}$

$\Rightarrow$ V = m$\frac{1}{T}$ where, m = slope of curve.

$\Rightarrow$ VT = constant

$\therefore$ TdV + VdT = 0 ..... (i)

$\Rightarrow$ dV = $-\frac{V}{T}$dT

As, Q = $\Delta$U + W

CdT = C_VdT + pdV ..... (ii)

CdT = C_VdT $-$ $\frac{p}{T}$ VdT

$\Rightarrow$ C = C_V $-$ $\frac{pV}{T}$

= C_V $-$ R ($\because$ $\frac{pV}{T}$ = R)

For diatomic gas, C_V = $\frac{5}{2}$R

$\therefore$ C = C_V $-$ R = $\frac{5}{2}$R $-$ R = $\frac{3R}{2}$

Work done in the cyclic process

= Area of the loop ABCD

= (2p $-$ p) $\times$ (2V $-$ V) = pV

Conductivity ($\sigma$) of a metallic conductor is given by

$\sigma = {1 \over \rho } = {{n{e^2}\tau } \over m}$

where, $\rho$ = resistivity of conductor,

$\tau$ = relaxation time,

n = number of free electrons

and m = mass.

For metals, as the temperature increases, the relaxation time $\tau$ decreases because collisions become more frequent. Hence, the conductivity of a metallic conductor decreases with the increases of temperature.

As we know,

${C_v} = {n \over 2}R$ ...... (i)

${C_p} = \left( {{n \over 2} + 1} \right)R$ ..... (ii)

On dividing Eq. (ii) by Eq. (i), we get

${{{C_p}} \over {{C_v}}} = \gamma = {{\left( {{n \over 2} + 1} \right)} \over {{n \over 2}}} = 1 + {2 \over n}$

$ \Rightarrow {{{C_p}} \over {{C_v}}} = 1 + {2 \over n}$

Let E' be the new radiant energy.

According to Stefan's law,

E $\propto$ T⁴

$\Rightarrow$ ${{{E_2}} \over {{E_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$

$\Rightarrow$ ${{E'} \over E} = {\left\{ {{{(T/4)} \over T}} \right\}^4} = {\left( {{1 \over 4}} \right)^4}$

or, $E' = {E \over {256}}$

In adiabatic process,

pV^$\gamma$ = K

where, K is a constant.

Given that, $\gamma = {4 \over 3} \Rightarrow p{V^{4/3}} = K$

Taking logarithm on both sides, we get

$\log p + {4 \over 3}\log V = \log K$

$ \Rightarrow {{\Delta p} \over p} + {4 \over 3}{{\Delta V} \over V} = 0$

$ \Rightarrow {{\Delta V} \over V} = - {3 \over 4}{{\Delta p} \over p}$

Now, % change in volume,

${{\Delta V} \over V} \times 100 = - {3 \over 4}\left( {{{\Delta p} \over p} \times 100} \right)$

It is given that, pressure decreases by ${3 \over 4}$%,

i.e. ${{\Delta p} \over p} \times 100 = - {3 \over 4}$

$\therefore$ ${{\Delta V} \over V} \times 100 = - {3 \over 4} \times \left( { - {3 \over 4}} \right) = {9 \over {16}}$%

Hence, the volume increases by ${9 \over {16}}$%.

Efficiency of Carnot engine is given by

$\eta = 1 - {{{T_2}} \over {{T_1}}}$

Given, T₁ = 600 K

T₂ = 300 K

T'₁ = 1600 K and T'₂ = x K

Carnot engine has same efficiency between 600 K to 300 K and 1600 K to x K, so

$\left( {1 - {{{T_2}} \over {{T_1}}}} \right) = \left( {1 - {{T{'_2}} \over {T{'_1}}}} \right)$

Substituting all the given values in above equation, we get

$1 - {{300} \over {600}} = 1 - {x \over {1600}}$

$1 - {1 \over 2} = 1 - {x \over {1600}}$

${x \over {1600}} = 1 - {1 \over 2} = {1 \over 2}$

$ \Rightarrow x = {{1600} \over 2} = 800$ K

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}}$