Homework Help: What is the rms speed of each helium atom...

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
What is the rms speed of each helium atom?
Answer in units of m/s.

Sol:

The average kinetic energy is given by:

KE_atom = (PV)/N

Number of atoms (N)= Avogadro's number (A) * n
where N = A*n = A*(PV/RT)

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

KE_atom = (96235*0.01414 /3.456 x 10^23 )J/atom
      = 3.94 X 10^-21 J/atom

KE_a = (1/2) m v^2
m = (Molecular Mass (Mm) / A)
m = (0.004 kg/mole / 6.02214 x 10^23 atoms/mole)
m = 6.42 x 10^-27 kg/atom
v^2 = (2 KE_a)/m
v = sqrt((2 KE_a) / m)
v = sqrt((2 * 3.94 X 10^-21 J/atom) / 6.42 x 10^-27 kg/atom)
v = 1108 m/s

Homework Help: What is the average kinetic energy of each helium atom...

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
What is the average kinetic energy of each helium atom?
Answer in units of J.

Sol:

The average kinetic energy is given by:

KE_atom = (PV)/N

Number of atoms (N)= Avogadro's number (A) * n
where N = A*n = A*(PV/RT)

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

KE_atom = (96235*0.01414 /3.456 x 10^23 )J/atom
      = 3.94 X 10^-21 J/atom

Chemistry Homework Help: How many atoms of helium gas are required to fill a balloon to diameter..

Q)The molecular mass of helium is 4 g/mol, the Boltzmann’s constant is 1.38066 ×
10−23 J/K, the universal gas constant is 8.31451 J/K · mol, and Avogadro’s number
is 6.02214 × 1023 1/mol. Given: 1 atm = 101300 Pa.
How many atoms of helium gas are required to fill a balloon to diameter 30 cm at 12◦C and 0.95 atm?

Sol:

PV = nRT
n = PV/RT
Number of atoms (N)= Avogadro's number (A) * n
N = A*n = A*(PV/RT)
P = 0.95 * 101300 Pa
P = 96235 Pa
V = (4/3) pi R^3
R = 0.15 m
V = (4/3) pi (0.15 m)^3
V = 0.01414 m^3
T = 273.15 + 12
T = 285.15 K

N = 6.02214 x 10^23 atoms/mole * (96235 Pa * 0.01414 m^3) / (8.31451 J/K-mol * 285.15 K)
N = 3.456 x 10^23 atoms

Networking Assignment Help: Explain the RSA Public Key Encryption Scheme in detail...

Q)Explain the RSA Public Key Encryption Scheme in detail.In particular ,how do the encryption and decryption algorithms work? Explain also how you choose the public and private keys.Suppose a user picks n=17*5; i.e .,p=17 and q=5.How should he pick the public key and the private key (i.e., d and e)? Give the encryption and decryption algorithms.

Sol:

RSA Public Key Encryption which was invented by Cocks (GCHQ), independently, by Rivest, Shamir and Adleman (MIT). Say for example let ‘p’ and ‘q’ is the two large prime numbers.

Let N = p*q be the modulus, then Choose ‘e’ relatively prime to (p-1)*(q-1)

Find d such that e*d = 1 mod (p-1)*(q-1)

Here the Public key is given by (N, e) and the Private Key is given by ‘d’.

The encryption as well as the decryption algorithm works as follows:

To encrypt the message M compute

C = M^e mod N

And to decrypt the message C compute

M = C^d mod N

Please recall that e and N are public, hence if attacker can factor N, then he can definitely use ‘e’ to easily find the value of ‘d’ since e*d = 1 mod (p-1)*(q-1)

Factoring the modulus will break the RSA, therefore it is not known that whether the factoring is the only one way to break RSA.

Suppose a user picks N=17*5; i.e., p=17 and q=5.

Hence N= 85,

(P-1)*(q-1) = 16*4=64

Choose ‘e’ such that it is relatively prime to 64. Hence e=5

Find‘d’ such that e*d = 1 mod 64, we find that d = 7 works

Public key: (N, e) = (85, 5)

Private Key: d = 7

Encryption as well as decryption algorithm works as follows:

Public key: (N, e) = (85, 5)

Private Key: d = 7

Suppose message M = 8

Cipher text C is computed as

C = M^e mod N = 8⁵ = 32768 = 17 mod 85

Decrypt C to recover the message M by

M = C^d mod N = 17⁷ = 410,338,673

= 12,434,505 * 85 + 8 = 8 mod 85

Homework Help: What is the speed of the block when the spring has been compressed...

A block with a mass of 0.148kg is traveling at 8.4m/s on a horizontal,frictionless surface when it hits a spring with a spring constant of 722 N/m. What is the compression of the spring when the block has slowed to 3.9 m/s?What is the speed of the block when the spring has been compressed by 0.049m?

Looking for an answer to this question, then comment on this post!!!

Physics Assignment Help: How much work is done moving a...

Q)How much work is done moving a 2.5 kg book to a shelf 3.0 m high?

a.What is the potential energy of the book in #7 above?
b.If the book in #7 above, falls off the shelf, how much kinetic energy will it have when it hits the floor

Sol:

Work done = Force*displacement

Force=m*g=2.5*9.8=24.5N

Work done= 24.5*3=73.5Joules

a) Potential energy = work done= m*g*h= 73.5Joules

b) Kinetic energy when it hits the floor will be at t=0,

V=U + at=0+9.8*0=0

KE= ½*m*v² = 0J

Chemistry Assignment Help: What is the equilibrium concentration of H2S...

Q)Kc = 2.6 × 10^8 at 825 K for the reaction

2H2(g) + S2(g) <===> 2H2S(g)

The equilibrium concentration of H2 is 0.0020 M and that of S2 is 0.0010 M. What is the equilibrium concentration of H2S?

1. 0.0010 M
2. 1.02 M
3. 10 M
4. 0.10 M

Sol:

K_c = [H2S]²/[H2]²*[S2]

2.6*10⁸ = [H2S]²/[0.002]²[0.001]

Hence equilibrium concentration of [H2S]= 1.02M