Thermodynamics

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New answer posted•a year ago

Thermodynamics Class 11th

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?

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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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New answer posted•a year ago

Thermodynamics Class 11th

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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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New answer posted•a year ago

Thermodynamics Class 11th

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________.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

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

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

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

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

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.

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

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

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$r_{α} = \frac{m_{α} v}{q_{α} B}$

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

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

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

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

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.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

85gm NH₃ = 5 moles of NH₃

Enthalpy change for 1 mol = 23.4 kJ

Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

T₁ = 227°C &n

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter One

2.4g coal is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure. The temperature of the calorimeter rises from 298 K to 300 K. The enthalpy change during the combustion of coal is -x kJ mol^-1. The value of x is ____________.

(Given : Heat capacity of bomb calorimeter 20.0 kJ K^-1. Assume coal to be pure carbon)

10 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$C (s) + O_{2} (g) \to C O_{2} (g) Δ n g = 0$

$q_{C} = 20 \times 2 = 40 k J$

Number of moles of C = $\frac{2.4}{12} = 0.2 m o l e s$

$Δ H_{C} = \frac{40 k J}{0.2 m o l e} = 200 k J / m o l e$

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

For the reaction

H₂F₂(g) ® H₂(g) + F₂(g)

$Δ U = - 59.6 k J m o l^{- 1} a t 27 ° C$

The enthalpy change for the above reaction is ()__________kJ mol^-1. [nearest integer]

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$Δ H = Δ U + Δ n_{g} R T$

$Δ H = - 59.6 + 1 \times 8.314 \times 300 \times 1 0^{- 3} = - 57.10 k J / m o l$

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