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2 months 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

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Payal Gupta

Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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2 months 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________.

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Vishal Baghel

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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2 months 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___________.

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Vishal Baghel

Contributor-Level 10

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

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2 months 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.

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Payal Gupta

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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2 months 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

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Payal Gupta

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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2 months 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.

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Vishal Baghel

Contributor-Level 10

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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2 months 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.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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2 months 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)

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Vishal Baghel

Contributor-Level 10

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

$q_{C} = 20 * 2 = 40 k J$

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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2 months 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]

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alok kumar singh

Contributor-Level 10

$Δ H = Δ U + Δ n_{g} R T$

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

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2 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

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alok kumar singh

Contributor-Level 10

44.4 litres of He $\frac{44.8}{22.4} m o l e s$

$η = 2$ moles of He

Q = $η C_{v} Δ T$ (for fixed capacity (v = constant)

= $2 * \frac{R}{γ - 1} Δ T$

= $\frac{2 * R}{\frac{5}{3} - 1} * 20$

= $= 2 * \frac{3 R}{2} * 20$

Q = 60R = 60 * 8.3

Q = 498 J

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