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3 months ago

Class 11th Equilibrium

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_P for the reaction is

[Given R = 0.082 L atm K^-¹ mol^-¹]

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alok kumar singh

Contributor-Level 10

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5

...more

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5 atm

$K_{P} = \frac{1.5 * 1.5}{1} = 2.25$

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3 months ago

Class 11th Chemistry Chemical Bonding and Molecular Structure

Consider the species CH₄, $N H_{4}^{+} a n d B H_{4}^{-} .$ Choose the correct option with respect to the there species

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alok kumar singh

Contributor-Level 10

Species = CH₄ $N H_{4}^{+}$ $B H_{4}^{?}$

Electrons = 10 10

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Class 11th Physics Thermodynamics

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

In case of adiabatic process -

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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Class 11th Physics

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be:

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Raj Pandey

Contributor-Level 9

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 * \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 * 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 * 12 * 3.14 * 1 0^{- 4} = 2.8 * 1 0^{- 4} J$

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Class 11th Physics

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

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Raj Pandey

Contributor-Level 9

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4 r_{B}^{3}$

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Class 11th Physics

Match List – I with List – II

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

Use formula for M.I.

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Class 11th Physics Motion in Plane

Motion of a particle in x – y plane is described by a set of following equations x = 4sin $(\frac{π}{2} - ω t)$ m and y = 4sin $(ω t) m .$ The path of a particle will be:

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Raj Pandey

Contributor-Level 9

$x = 4 s i n (\frac{π}{2} - ω t)$ - (i)

$y = 4 s i n (ω t)$ - (ii)

From (i) and (ii) cos^{2
$ω t + s i n^{2} ω t = {(\frac{x}{4})}^{2} + {(\frac{y}{4})}^{2} = 1$}

$\Rightarrow x^{2} + y^{2} = {(4)}^{2}$

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Class 11th Structure of Atom

Which of the following is the correct plot for the probability density $ψ^{2} (r)$ as a function of distance 'r' of the electron from the nucleus for 2s orbital?

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alok kumar singh

Contributor-Level 10

For 2S, number of radial moles = $n - l - 1 = 2 - 0 - 1 = 1$ and $Ψ^{2} (r)$ will always be positive

$∴ o p t i o n (B) i s c o r r e c t ?$ it has one radial node and it is positive.

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Class 11th Work, Energy and Power

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

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Raj Pandey

Contributor-Level 9

a = k2rt2

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

? tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note ® Power delivered by centripetal force will be zero.

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3 months ago

Class 11th Chemistry Some Basic Concepts of Chemistry

Using the rules for significant figures, the correct answer for the expression ^{$\frac{0.02858 * 0.112}{0.5702}$}will be

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

$\frac{0.02858 * 0.112}{0.5702} = 0.0561$

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