Question

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at 50∘C has a total translational kinetic energy of 4000 J.A) Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C is:________a) (16)(2000m/s)=32000m/sb) (4)(2000m/s)=8000m/sc) 2000m/sd) (14)(2000m/s)=500m/se) (116)(2000m/s)=125m/sf) none of the aboveB) The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50∘C is:________a) (16)(4000J)=64000Jb) (4)(4000J)=16000Jc) 4000Jd) (14)(4000J)=1000Je) (116)(4000J)=150Jf) none of the aboveC) The temperature of the diatomic hydrogen gas sample is increased to 100∘C. The root-mean-square speed vrms for diatomic hydrogen at 100∘C is:______a) (2)(2000m/s)=4000m/sb) (2√)(2000m/s)=2800m/sc) 2000m/sd) (12√)(2000m/s)=1400m/se) (12)(2000m/s)=1000m/sf) none of the above

Answer 1

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s