Heat and Thermodynamics

$\because$ dW = PdV

and there is no change in volume in isochoric process.

$\therefore$ dW = 0

Adiabatic process is used as a correction in the determination of the speed of sound in an ideal gas. The p-V diagram of this process is shown in (Q). The work done on the gas in an adiabatic process is given by ${W_{1 \to 2}} = {{{p_2}{V_2} - {p_1}{V_1}} \over {\gamma - 1}}$.

Compare the given relation, $\Delta$U = $\Delta$Q $-$ p$\Delta$V, with the first law of thermodynamics, $\Delta$Q = $\Delta$U + $\Delta$W, to get work done by the gas as $\Delta$W = p$\Delta$V. This is the work done in an isobaric process. The p-V diagram of this process is shown in (P). The work done in this process is ${W_{1 \to 2}} = \int {pdV = p{V_1} - p{V_2}} $.

Initially both the compartments has same pressure as they are in equilibrium.

Suppose spring is compressed by x on heating the gas.

Let A be the area of cross-section of piston. As gas is ideal monoatomic, so

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}{V_2}} \over {{T_2}}}$ ...... (i)

Force on spring by gas = kx

$\therefore$ ${P_2} = {P_1} + {{kx} \over A}$ ...... (ii)

Case I : When ${V_2} = 2{V_1}$, ${T_2} = 3{T_1}$

From Eqn. (i)

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(2{V_1})} \over {3{T_1}}} \Rightarrow {P_2} = {3 \over 2}{P_1}$

Putting this value in eqn. (ii) we get

${3 \over 2}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 2}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1} - {V_1}} \over A} = {{{V_1}} \over A}$

Energy stored in the spring

$ = {1 \over 2}k{x^2} = {1 \over 2}(kx)(x) = {{{P_1}{V_1}} \over 4}$

So, option (a) is correct.

Change in internal energy,

$\Delta U = {f \over 2}({P_2}{V_2} - {P_1}{V_1}) = {3 \over 2}\left( {{3 \over 2}{P_1} \times 2{V_1} - {P_1}{V_1}} \right) = 3{P_1}{V_1}$

So, option (b) is correct.

Case II : When ${V_2} = 3{V_1}$ and ${T_2} = 4{T_1}$

From Eqn. (i),

${{{P_1}{V_1}} \over {{T_1}}} = {{{P_2}(3{V_1})} \over {4{T_1}}} \Rightarrow {P_2} = {4 \over 3}{P_1}$

$x = {{{V_2} - {V_1}} \over A} = {{2{V_1}} \over A}$

From eqn. (ii),

${4 \over 3}{P_1} = {P_1} + {{kx} \over A} \Rightarrow kx = {{{P_1}A} \over 3}$

Gas is heated very slowly so pressure on the other compartment remains same.

Work done by gas = Work done by gas on atmosphere + Energy stored in spring.

${W_g} = {P_1}Ax + {1 \over 2}k{x^2} = {P_1}(2{V_1}) + {1 \over 2}\left( {{{{P_1}A} \over 3}} \right)\left( {{{2{V_1}} \over A}} \right)$

$ = 2{P_1}{V_1} + {1 \over 3}{P_1}{V_1} = {7 \over 3}{P_1}{V_1}$

So, option (c) is correct.

Heat supplied to the gas,

$\Delta Q = {W_g} + \Delta U$

$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}({P_2}{V_2} - {P_1}{V_1})$

$ = {7 \over 3}{P_1}{V_1} + {3 \over 2}\left( {{4 \over 3}{P_1} \times 3{V_1} - {P_1}{V_1}} \right)$

$ = {7 \over 3}{P_1}{V_1} + {9 \over 2}{P_1}{V_1} = {{41} \over 6}{P_1}{V_1}$

So, option (d) is incorrect.

The internal energy of one mole of an ideal gas at temperature T is given by $U = {f \over 2}RT$, where f is the degrees of freedom of the gas molecule. The degress of freedom of the gas molecule. The degrees of freedom for hydrogen (diatomic) and helium (monatomic) gases are f_H2 = 5 and f_He = 3, respectively. Thus, ${U_{{H_2}}} = {5 \over 2}RT$ and ${U_{He}} = {3 \over 2}RT$. The total internal energy of the gas mixture is

${U_{total}} = {U_{{H_2}}} + {U_{He}} = {5 \over 2}RT + {3 \over 2}RT = 4RT$.

The mixture contains two moles of the gases. The internal energy per mole of the mixture is ${U_{mix}} = {U_{total}}/2 = 2RT$.

The specific heat at constant volume is given by ${C_v} = dU/dT$. Thus, the specific heats at constant volume for helium and the mixture are

${C_{v,He}} = d{U_{He}}/dT = {3 \over 2}R$, and

${C_{v,mix}} = d{U_{mix}}/dT = 2R$

The specific heats at constant pressure, ${C_p} = {C_v} + R$, for these gases are

${C_{p,He}} = {C_{v,He}} + R = {5 \over 2}R$, and

${C_{p,mix}} = {C_{v,mix}} + R = 3R$.

The ratio of specific heats, $\gamma = {C_p}/{C_v}$, are ${\gamma _{He}} = 5/3$ and ${\gamma _{mix}} = 3/2$. The speed of sound, in a gas of molecular mass M, is given by ${v_s} = \sqrt {\gamma RT/M} $. The molecular mass of the gas mixture is

${M_{mix}} = {{{n_{{H_2}}}{M_{{H_2}}} + {n_{He}}{M_{He}}} \over {{n_{{H_2}}} + {n_{He}}}}$

$ = {{(1)(2) + (1)(4)} \over {1 + 1}} = 3$ g/mol,

where n_H2 = 1 and n_He = 1 are the number of moles of hydrogen and helium in the gas mixture. The ratio of the speeds of sound in the gas mixture and helium is

${{{v_{s,mix}}} \over {{v_{s,He}}}} = {{\sqrt {{\gamma _{mix}}RT/{M_{mix}}} } \over {\sqrt {{\gamma _{He}}RT/{M_{He}}} }} = \sqrt {{{{\gamma _{mix}}} \over {{\gamma _{He}}}}{{{M_{He}}} \over {{M_{mix}}}}} $

$ = \sqrt {{{(3/2)(4)} \over {(5/3)(3)}}} = \sqrt {{6 \over 5}} $

The rms speed of the atoms/molecules is given by ${v_{rms}} = \sqrt {3RT/M} $. The ratio of the rms speed of helium atoms to that of hydrogen molecules is

${{{v_{rms,He}}} \over {{v_{rms,{H_2}}}}} = {{\sqrt {3RT/{M_{He}}} } \over {\sqrt {3RT/{M_{{H_2}}}} }} = \sqrt {{{{M_{{H_2}}}} \over {{M_{He}}}}} = \sqrt {{2 \over 4}} = \sqrt {{1 \over 2}} $.

If the junction temperature is $T,$ then

${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$

${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$

$ \Rightarrow 200 - 2T = 2T + T$

$ \Rightarrow T = {40^ \circ }C$

$\therefore$ ${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$

where, ${C_v} = $ molar specific heat at constant volume.
For $BC,$ $\Delta T = - 200\,K$
$\therefore$ $\Delta {U_{BC}} = - 500R$

Effective area of incident light = $\pi$r²

Total energy per second = 912 $\times$ $\pi$r²

From Stefan's law, we have

E = Ae $\sigma$ (T⁴ $-$ T$_0^4$)

For black body emissivity, e = 1.

A = 4$\pi$r²

912 $\times$ $\pi$r² = 4$\pi$r² $\times$ 5.7 $\times$ 10^$-$8 [T⁴ $-$ 300⁴]

$({T^4} - {300^4}) = {{912} \over {4 \times 5.7 \times {{10}^{ - 8}}}} = 40 \times {10^8}$

${T^4} = 40 \times {10^8} + {3^4} \times {10^8}$

${T^4} = (90 + 81) \times {10^8} \Rightarrow 121 \times {10^8}$

$T = {(121)^{1/4}} \times 100 = \sqrt {11} \times 100 \Rightarrow 3.3 \times 100$

T = 330 K

T is final temperature, when equilibrium is reached

Heat gained = Heat lost

$n{C_p}\Delta T = n{C_v}\Delta T$

$2 \times {{7R} \over 2} \times (T - 400) = 2 \times {{3R} \over 2} \times (700 - T)$

$7T - 2800 = 2100 - 3T$

$10T = 4900$

$T = 490K$

Heat lost by monoatomic gas = Heat gained by diatomic gas

$2 \times {5 \over 2}R \times (700 - T) = 2 \times {5 \over 2}R(T - 400)$

$3500 - 5T = 7T - 2800$

$12T = 6300$ or, $T = 525K$

Work done by monoatomic gas:

$ = {n_1}R\Delta T$ ($\because$ PV = nRT)

$ = 2 \times R \times (700 - 525)$ ($\Delta$T is negative)

$ = - 350R$

Work done by diatomic gas $ = {n_2}R\Delta T$

$ = 2R(525 - 400) = 250R$. (There is gain in temperature)

The net work done = $-$350 + 250 = $-$100R and $\Delta$T is positive.

According to definition of specific heat capacity,

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

$ \Rightarrow \Delta Q = mC\Delta T$ $\therefore$ ${{\Delta Q} \over {\Delta t}} = mC{{\Delta T} \over {\Delta t}}$

Rate of heat absorbed $R = {{\Delta Q} \over {\Delta t}}$

$ \Rightarrow {{\Delta Q} \over {\Delta t}} \propto C$

(a) In 0-100 K,

C increases with T but not linearly. So R increases but not linearly.

(b) As $\Delta Q = mC\Delta T$

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

= m area under C-T curve

From the graph it is clear that area under C-T is more in 400-500 K than in 0-100 K.

Therefore, heat absorbed in 0-100 K is less than in 400-500 K.

(c) In 400-500 K,

C remains constant so there is no change in R.

(d) In 200-300 K,

C increases so R increases.

From the given figure, it can be concluded that process FG is isothermal and process FH is adiabatic.

${\gamma _{mono}} = 1 + {2 \over f} = {5 \over 3}$

${P_0}V_G^{5/3} = 32{P_0}V_0^{5/3}$

${V_G} = {(32)^{3/5}}{V_0} = 8{V_0}$

$\Delta {W_{GE}} = {P_0}({V_0} - 32{V_0}) = - 31{P_0}{V_0}$

Thus, the correct mapping is (P) $\to$ (4).

$\Delta {W_{GH}} = {P_0}(8{V_0} - 32{V_0}) = - 24{P_0}{V_0}$

Thus, the correct mapping is (Q) $\to$ (3).

$\Delta {W_{FH}} = {{{P_0}(8{V_0}) - 32{P_0}{V_0}} \over {1 - (5/3)}} = {{ - 24{P_0}{V_0}} \over { - (2/3)}} = 36{P_0}{V_0}$

Thus the correct mapping is (R) $\to$ (2).

$\Delta {W_{FG}} = 32R{T_0}\ln 32 = 32R{T_0}\ln {2^5} = 160R{T_0}\ln 2 = 160{P_0}{V_0}\ln 2$

Thus, the correct mapping is (S) $\to$ (1).

To determine the ratio of the densities of the two non-reactive monoatomic ideal gases, we can use the ideal gas law and the given information. The ideal gas law is given by:

$ PV = nRT $

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the gas constant
• T is the temperature

Given that the gases are monoatomic and non-reactive, their molecular masses are directly proportional to their atomic masses. Let the atomic masses of the two gases be $M_1$ and $M_2$, with $M_1 : M_2 = 2 : 3$.

The density ($\rho$) of a gas is given by the formula:

$ \rho = \frac{m}{V} $

Where $m$ is the mass and $V$ is the volume. Using the ideal gas law, we can relate density to pressure and temperature.

From the ideal gas law:

$ PV = nRT $

Since $ n = \frac{m}{M} $, where $M$ is the molar mass, we have:

$ P = \frac{mRT}{MV} $

Rearranging to solve for density ($\rho\ = \frac{m}{V}$):

$ \rho = \frac{PM}{RT} $

Thus, the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to the temperature. Given that the ratio of their partial pressures $P_1 : P_2 = 4 : 3$, we can write:

$ \rho_1 = \frac{P_1 M_1}{RT} $

$ \rho_2 = \frac{P_2 M_2}{RT} $

Taking the ratio of the densities:

$ \frac{\rho_1}{\rho_2} = \frac{\left(\frac{P_1 M_1}{RT}\right)}{\left(\frac{P_2 M_2}{RT}\right)} = \frac{P_1 M_1}{P_2 M_2} $

Substituting the given ratios, $ P_1 : P_2 = 4 : 3 $ and $ M_1 : M_2 = 2 : 3 $, into the equation, we get:

$ \frac{\rho_1}{\rho_2} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9} $

Therefore, the ratio of their densities is $ \frac{8}{9} $, which corresponds to option D. Consequently, the answer is:

Option D: 8 : 9

Let L and A be length and area of cross-section of each block.

$\therefore$ Thermal resistance of block 1 is

$\therefore$ ${R_1} = {L \over {\kappa A}}$

Thermal resistance of block 2 is

${R_2} = {L \over {2\kappa A}}$

In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is

${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$ ...... (i)

$\therefore$ Rate of heat flow in configuration I is

$\therefore$ ${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$ ..... (ii)

In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is

${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$ ...... (iii)

${R_P} = {1 \over 3}{L \over {\kappa A}}$

$\therefore$ Rate of same amount of heat flow in configuration II is

$\therefore$ ${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$ ...... (iv)

Divide (ii) by (iv), we get

${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} \times {{2\kappa A} \over {3L}} = {2 \over 9}$ (Using (i) and (iii))

$t' = {2 \over 9}t = {2 \over 9} \times 9s = 2s$

Here's a breakdown of how to solve this problem:

Understanding the Concepts

• Ideal Gas: An ideal gas is a theoretical gas that follows the ideal gas law perfectly. The ideal gas law is a relationship between pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$) of a gas:

$PV = nRT$, where $R$ is the ideal gas constant.

• Heat Capacity: Heat capacity is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius (or 1 Kelvin). For an ideal gas, we can use the specific heat capacity at constant volume ($C_v$) or at constant pressure ($C_p$).


• Constant Volume Process: In a constant volume process, the volume of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_v\Delta T$

• Constant Pressure Process: In a constant pressure process, the pressure of the gas remains constant. The heat required to raise the temperature is given by:

$Q = nC_p\Delta T$

• Molar Heat Capacities of Helium: Helium is a monatomic gas. For monatomic gases, we have:


• $C_v = \frac{3}{2}R$


• $C_p = \frac{5}{2}R$

Solving the Problem

1. Identify the Process: Since the balloon is fully expandable, the pressure remains constant. This is a constant pressure process.


2. Calculate the Heat: Use the formula for heat at constant pressure:

$Q = nC_p\Delta T$

1. Substitute the Values:

• $n = 2 \text{ moles}$


• $C_p = \frac{5}{2}R = \frac{5}{2}(8.31 \text{ J/mol.K})$


• $\Delta T = 35^\circ C - 30^\circ C = 5 K$

1. Solve for Q:

$Q = (2 \text{ moles}) \times \left(\frac{5}{2} \times 8.31 \text{ J/mol.K}\right) \times (5 K) $

$Q = 207.75 \text{ J}$

Answer:

The amount of heat required to raise the temperature of the helium gas is approximately 208 J. Therefore, the correct answer is Option D.

The root mean square (rms) speed of a gas is given by the formula :

$v_{rms} = \sqrt{\frac{3kT}{m}}$

where :

• $k$ is the Boltzmann constant
• $T$ is the temperature in Kelvin
• $m$ is the mass of one molecule of the gas

Given that the atomic mass of helium is 4 amu and that of argon is 40 amu, we can find the rms speed ratio by comparing the rms speeds of helium and argon, as temperature $T$ and Boltzmann constant $k$ are the same for both gases :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{m_{\text{argon}}}{m_{\text{helium}}}}$

Substituting the masses :

$m_{\text{helium}} = 4 \text{ amu}$

$m_{\text{argon}} = 40 \text{ amu}$

we get :

$\frac{v_{rms} \left( \text{helium} \right)}{v_{rms} \left( \text{argon} \right)} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$

Thus, the ratio of the rms speeds of helium to argon is approximately 3.16.

The correct answer is :

Option D : 3.16

Let the steady state temperature of the middle plate to T₀. In the steady state, the heat radiated per unit time by the middle plate is equal to the heat received per unit time by it. The middle plate radiates heat from both the surfaces and receives the heat radiated by the first and the third plates.

Stefan-Boltzmann law gives the rate of heat loss and heat gain by the middle plate as

$d{Q_{out}}/dt = \sigma AT_0^4 + \sigma AT_0^4$,

$d{Q_{in}}/dt = \sigma A{(2T)^4} + \sigma A{(3T)^4}$.

The steady state condition, $d{Q_{out}}/dt = d{Q_{in}}/dt$

$\sigma A{(2T)^4} + \sigma A{(3T)^4} = \sigma 2A{({T_0})^4}$

$16{T^4} + 81{T^4} = 2{({T_0})^4}$

$97{T^4} = 2{({T_0})^4}$

${({T_0})^4} = {{97} \over 2}{T^4} = {\left( {{{97} \over 2}} \right)^{1/4}}T$

Initially

V₁ = 5.6 l, T₁ = 273 K, P₁ = 1 atm,

$\gamma = {5 \over 3}$ (For monoatomic gas)

The number of moles of gas is $n = {{5.6l} \over {22.4l}} = {1 \over 4}$

Finally (after adiabatic compression)

V₂ = 0.7 l

For adiabatic compression ${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}$

$\therefore$ ${T_2} = {T_1}{\left( {{{{V_1}} \over {{V_2}}}} \right)^{\gamma - 1}} = {T_1}{\left( {{{5.6} \over {0.7}}} \right)^{{5 \over 3} - 1}} = {T_1}{(8)^{2/3}} = 4{T_1}$

Work done during an adiabatic process is

$W = {{nR[{T_1} - {T_2}]} \over {(\gamma - 1)}} = {{{1 \over 4}R[{T_1} - 4{T_1}]} \over {\left[ {{5 \over 3} - 1} \right]}} = - {9 \over 8}R{T_1}$

Negative sign shows that work is done on the gas.