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An ideal gas undergoes isothermal expansion at constant pressure.  During the process :
Option: 1  enthalpy increases but entropy decreases.
Option: 2  enthalpy remains constant but entropy increases.
Option: 3  enthalpy decreases but entropy increases.
Option: 4  Both enthalpy and entropy remain constant.  

\Delta H = nC_{p}\Delta T \\ \Delta S = nR ln (V_{f}/V_{i})\geqslant 0

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Posted by

vishal kumar

 


The major product obtained in the following reaction is : 
Option: 1


Option: 2

Option: 3 

Option: 4 

The major product is represented by the option (D). DIBAL−H reduces esters to aldehydes. It also reduces carboxylic acids to aldehydes. 

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Posted by

vishal kumar

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A plane is inclined at an angle \alpha =30^{0} with respect to the horizontal. A particle is projected with a speed u=2ms^{-1}from the base of the plane , making an angle\theta =15^{0} With respect to the plane as shown in the  figure . The distance from the base ,at which the particle hits the plane is close to:\left ( Take \, \, g=10ms^{-2} \right )
Option: 1 20cm
Option: 2 18cm
Option: 3 26cm
Option: 4 14cm

As we know on inclined plane the range is given by

R=\frac{2u^{2}}{g}\frac{\sin \beta \cos \left ( \beta +\alpha \right )}{\cos ^{2}\alpha }

\beta =15^{0}\, \, ;\alpha =30^{0}\, \, ;u=2m/s

R=\frac{2\times 2^{2}\sin 15^{0}\cos \left ( 45 \right )}{10\times \cos \left ( 30^{0} \right )}

     = 20 cm \, \, approx

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Posted by

Ritika Jonwal

The dimension of stopping potential V_0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :   
Option: 1 h^{-2/3}c^{-1/3}G^{4/3}A^{-1}
 
Option: 2 [h]^0[G]^{-1}[c]^5[A]^{-1}

Option: 3 h^{2}G^{3/2}c^{1/3}A^{-1}  

Option: 4 h^{1/3}G^{2/3}c^{1/3}A^{-1}
 

 

 

 Let V_0=[h]^p[G]^q[c]^r[A]^s

Now, K_{max}=eV_0\Rightarrow [K_{max}]=[eV_0]\Rightarrow [V_0]=\frac{[K_{max}]}{[e]}\Rightarrow [V_0]=\frac{[ML^2T^{-2}]}{[AT]}\therefore [V_0]=[ML^2T^{-3}A^{-1}]

E=hf\Rightarrow [h]=\frac{[E]}{[f]}\Rightarrow \frac{[ML^2T^{-2}]}{[T^{-1}]}\Rightarrow [h]=[ML^2T^{-1}]

 

F=\frac{Gm^2}{r^2}\Rightarrow [G]=\frac{[F][r^2]}{[m^2]}=\frac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]

So, [ML^2T^{-3}A^{-1}]=[ML^2T^{-1}]^p[M^{-1}L^3T^{-2}]^q[LT^{-1}]^r[A]^s

From here we will get: p-q=1---------(1)

2p+3q+r=2----------(2)

-p-2q-r=-3----------(3)

s=-1-----------(4)

From equation 1, 2,3 and 4 we will get: p=0,q=-1, r=5 and s=-1

So, [V_0]=[h]^0[G]^{-1}[c]^5[A]^{-1}

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Posted by

vishal kumar

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The number of chiral carbons present in the molecule given below is __________.
Option: 1 5
Option: 2 5
Option: 3 -
Option: 4 -
Option: 5 -
Option: 6 -
Option: 7 -
Option: 8 -

Total chiral carbon = 5

 

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Posted by

Kuldeep Maurya

Which one of the following graphs is not correct for ideal gas ? d = Density, P = Pressure, T = Temperature  
Option: 1 I
Option: 2 I
Option: 3 II
Option: 4 II
Option: 5 IV
Option: 6 IV
Option: 7 III
Option: 8 III

For ideal Gas 

\mathrm{d}=\frac{\mathrm{P} \times \mathrm{M}}{\mathrm{RT}}

‘II’ Graph is incorrect.

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Posted by

Kuldeep Maurya

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If AB4 molecules is a polar molecule, a possible geometry of AB4 is :
 
Option: 1 Square pyramidal
Option: 2 Square pyramidal
Option: 3 Tetrahedral
Option: 4 Tetrahedral
Option: 5 Rectangular planar
Option: 6 Rectangular planar
Option: 7 Square planar
Option: 8 Square planar

(1) If AB4 molecule is a square pyramidal then it has one lone pair and their structure should be

and it should be polar because dipole moment of lone pair of 'A' never be cancelled by others.

(2) If AB4 molecule is a tetrahedral then it has no lone pair and their structure should be

and it should be non polar due to perfect symmetry.


(3) If AB4 molecule is a rectangular planar then

it should be non-polar because the vector sum of dipole moment is zero.

(4) If AB4 molecule is a square planar then

it should be non-polar because the vector sum of the dipole moment is zero.

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Posted by

Kuldeep Maurya

Which of the following is used for the preparation of colloids ?
 
Option: 1 Ostwald Process
Option: 2 Ostwald Process
Option: 3 Van Arkel Method
Option: 4 Van Arkel Method
Option: 5 Bredig's Arc Method  
Option: 6 Bredig's Arc Method  
Option: 7 Mond Process
Option: 8 Mond Process

Bredig's Arc method is used to form metal colloids.

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Posted by

Kuldeep Maurya

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On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with H_{2} in the presence of a catalyst gives another gas (C) which is basic in nature. (A) should not be :  
Option: 1 NaN_{3}
Option: 2 NaN_{3}
Option: 3 Pb(NO_{3})_{2}
Option: 4 Pb(NO_{3})_{2}
Option: 5 (NH_{4})_{2}Cr_{2}O_{7}
Option: 6 (NH_{4})_{2}Cr_{2}O_{7}
Option: 7 NH_{4}NO_{2}
Option: 8 NH_{4}NO_{2}

Basic gas (C) must be ammonia (NH3). It means (B) gas should be N2 which is formed by the heating of compound (A).

 

Pb(NO3)2 is not giving Nitrogen gas.

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Posted by

Kuldeep Maurya

The metal mainly used in devising photoelectric cells is :  
Option: 1 Na
 
Option: 2 Na
 
Option: 3 Li
Option: 4 Li
Option: 5 Rb  
Option: 6 Rb  
Option: 7 Cs
Option: 8 Cs

Cs used in photoelectric cell as it has least ionization energy.

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Posted by

Kuldeep Maurya

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