Thermodynamics

2O₃(g) $?$ 3O₂(g)

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-)________. J mol^?1. (Nearest integer)

[Given: In 1.35 = 0.3 and R=8.3 J K^-1 mol^-1]

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2O₃(g) -> 3O₂

t = 0 a moles 0

-0.5a mole +0.75a mole

At Eq. 0.5a mole 0.75a mole

Total moles at eq = 0.5a + 0.75 a = 1.25a

$P_{O_{3}} = x_{O_{3}} * P_{T} = \frac{0.5 a}{1.25 a} * 1 = \frac{2}{5} a t m$

$P_{O_{2}} = x_{O_{2}} * p_{T} = \frac{0.75 a}{1.25 a} * 1 = \frac{3}{5} a t m$

$K_{p} = \frac{{(P_{O_{2}})}_{e q}^{3}}{{(P_{O_{3}})}_{e q}^{3}} = \frac{{(\frac{3}{5})}^{3}}{{(2 / 5)}^{2}} = 1.35 a t m$

$Δ G^{°} = - R T l n K P$

$\begin{array}{l} = - 8.3 * 300 l n 1.35 \\ = - 747 J / m o l e \end{array}$

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A carnot engine takes 5000 kcal of heat from a reservoir at $727 ° C$ and gives heat to a sink at $127 ° C .$ The work done by the engine is

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T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 * 10³ k cal

$\Rightarrow \frac{600}{1000} = \frac{w}{5000}$

w = 12.6 * 10⁶ J

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A cornot engine whose heat sinks at 27°C has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

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0.25 = 1 - $\frac{T_{2}}{T_{1}}$ $\frac{_{}}{_{}}$

0.25 = 1 - $\frac{(27 + 273)}{T_{1}} \Rightarrow \frac{300}{T_{1}} = 1 - 0.25$ $\frac{}{_{}} \frac{}{_{}}$

$\Rightarrow \frac{300}{T_{1}} = 0.75$

$T_{1} = \frac{300 * 100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

$η_{2} = 100 % η_{1} + η_{1}$

$= 2 η_{1}$

= 0.25 * 2

= 0.50

0.50 = 1 - $\frac{T_{2}'}{T_{1}}$ $\frac{_{}}{_{}}$

$\Rightarrow \frac{T_{2}'}{T_{1}} = 0.50$

$T_{1}' = \frac{300}{0.5}$

$T_{1}' = 600 k$

$T_{1}' - T_{1} =$ (600 – 400) = 200 k or 200° C

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4 months ago

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4K. The ratio of molal elevation constant to molal depression constant (K_b/K_f) is________.

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Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

Contributor-Level 10

$? T_{b} = i K_{b} m$

$? T_{f} = i K_{f} m$

$? \frac{4}{4} = \frac{K_{b} * 1.5}{K_{f} * 4.5}$

$\frac{K_{b}}{K_{f}} = 3$

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4 months ago

Amongst SF₄, XeF₄, CF₄ and H₂O, the number of species with two lone pairs of electrons is___________.

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Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

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Water has only two lone pair and XeF₄ has two lone pair electron in opposite plane of the central atom.

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A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

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Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

Contributor-Level 10

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

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4 months ago

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

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Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

Contributor-Level 10

$r = \frac{m v}{q B}$

$r_{α} = \frac{m_{α} v}{q_{α} B}$

$r_{P} = \frac{m_{P} v}{q_{P} B}$

$\frac{r_{α}}{r_{P}} = \frac{m_{α} q_{P}}{m_{P} q_{α}} = \frac{m_{α}}{m_{P}} (\frac{q_{P}}{q_{α}})$

$= \frac{4 m}{m} (\frac{q}{2 q}) = 2 : 1$

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4 months ago

17.0g of NH₃ completely vapourises at -33.42°C and 1 bar pressure and the enthalpy change the process is 23.4 kJ mol^-1. The enthalpy change for the vapourisation of 85g of NH₃ under the same conditions is_____ kJ.

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