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3 months ago

Class 11th Thermodynamics

n mole of perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A->B : Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and Pressure changes from P₁ to P₂.

B->C : Isobaric compression at pressure P₂ to initial volume V₁.

C->A : Isobaric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is:

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alok kumar singh

Contributor-Level 10

$W_{A B} = n R T l n \frac{2 V_{1}}{V_{1}} = n R T l n 2 .$

$W_{B C} = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{1}{2} n R T$

[At B, 2P₂ V₁ = nRT]

$W_{C A} = 0$ [CA is isochoric process].

$W_{A B C A} = W_{A B} + W_{B C} + W_{C A} = n R T (l n (2) - \frac{1}{2})$

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Class 11th Physics System of Particles and Rotational Motion

Moment of inertia (M.I) of four bodies, having same mass and radius, and reported as:

I₁ = M.I. of thin circular ring about its diameter

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

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alok kumar singh

Contributor-Level 10

$l_{1} = \frac{1}{2} m r^{2}$

$\begin{array}{l} l_{2} = \frac{1}{2} m r^{2} \\ l_{3} = \frac{1}{2} m r^{2} \\ l_{4} = \frac{2}{5} m r^{2} \end{array}$

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Class 11th Physics

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The

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alok kumar singh

Contributor-Level 10

$m ω^{2} . \frac{2 d}{3} = \frac{G . m . 2 m}{d^{2}} \Rightarrow ω = \sqrt[]{\frac{3 G m}{d^{3}}}$

$T = \frac{2 π}{ω} = 2 π \sqrt[]{\frac{d^{3}}{3 G m}}$

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Class 11th Physics

Match List I with List II.

List I List II

(a) Isothermal &nb

...more

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alok kumar singh

Contributor-Level 10

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant

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Class 11th Physics Motion in Straight Line

If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

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alok kumar singh

Contributor-Level 10

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.

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Class 11th Physics Thermal Properties of Matter

Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is 'a'. The metal sheet is heated uniformly, bya small temperature $Δ T,$ so that its new temperature is T + $Δ T .$ Calculate the increase in the volume of the metal box.

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alok kumar singh

Contributor-Level 10

$γ = 3 α$

$Δ V = γ V Δ T = 3 α . a^{3} . Δ T$

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Class 11th Chemistry

For a chemical reaction A + B $?$ C + D

$(Δ_{r} H^{Θ} = 80 k J m o l^{- 1})$ the entropy change $Δ_{r} S^{Θ}$ depends on the temperature T (in K) as $Δ_{r} S^{Θ} = 2 T (J K^{- 1} m o l^{- 1}) .$

Minimum temperature at which it will become spontaneous is-------------K. (Integer)

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Payal Gupta

Contributor-Level 10

$Δ_{r} H^{o} = 80 k J m o l^{- 1}$

$Δ_{r} S^{o} = 2 T k J m o l^{- 1}$

For a reaction to be spontaneous;

$Δ_{r} S^{o} > \frac{Δ_{r} H^{o}}{T}$

$T > \frac{Δ_{r} H^{o}}{Δ_{r} S^{o}}$

$T > \frac{80 * 1 0^{3}}{2 T}$

$T^{2} > 40000 K$

$T > 200 K$

So, minimum T at which reaction will be spontaneous is 200 K.

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Class 11th Chemistry Some Basic Concepts of Chemistry

The number of significant figures in 50000.020 * 10^-3 is………………

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Payal Gupta

Contributor-Level 10

50000.020 * 10^-3

The significant figure in the given number is 8.

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Class 11th Chemical Equilibrium

An homogeneous ideal gaseous reaction $A B_{2 (g)} ? A_{(g)} + 2 B_{(g)}$ is carried out in a 25 litre flask at 27°C. The initial amount of AB₂ was 1 mole and the equilibrium pressure was 1.9atm. The value of K_p is x * 10^-2. The value of x is-----------. (Integer answer).

[R = 0.08206 dm³ atm K^-1 mol^-1]

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Payal Gupta

Contributor-Level 10

$A B_{2} (g) ? A (g) + 2 B (g)$

Initial1 mole -

At equilirbium 1-x mole x mole2x mole

V = 25 L and

T = 300 K.

At equilibrium, P = 1.9 atm

Total moles at equilibrium, n = 1 + 2x

V = 25 L

T = 300 K

Using, PV = nRT

1.9 * 25 = (1 + 2x) * 0.08206 * 300

x = 0.465

Now;Partial pressure of AB2 at equilibrium = $\frac{0.535}{1.93} * 1.9 a t m$

Using ; $K_{p} = \frac{P_{A} . P_{B}^{2}}{P_{A B_{2}}}$

$K_{p} = \frac{(\frac{0.465 * 1.9}{1.93}) {(\frac{2 * 0.465}{1.93} * 1.9)}^{2}}{\frac{0.535}{1.93} * 1.9}$ = 0.728

K_P = 0.73

K_p = 73 * 10^-2

So; x = 73

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Class 11th Chemistry

A certain gas obeys $P (V_{m} - b) = R T .$ The value of ${(\frac{\partial z}{\partial P})}_{T} i s \frac{x b}{R T} .$ The value of x is------------.

(Integer answer) (Z : compressibility factor)

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Payal Gupta

Contributor-Level 10

$P (V_{m} - b) = R T$

$\frac{P V_{m}}{R T} - \frac{P b}{R T} = 1$

$Z = 1 + \frac{P b}{R T}$

${(\frac{\partial z}{\partial P})}_{T} = \frac{b}{R T}$

So, comparing with $\frac{x b}{R T}$

x = 1

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