The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50?C is 2000

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.

Part A.

The total translational kinetic energy of 1.0 mole of diatomic oxygen at 50?C is:

Choose the correct total translational kinetic energy.

Choose the correct total translational kinetic energy.

none of the above

Part B:

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:

(16)(4000J)=64000J (4)(4000J)=16000J 4000J (14)(4000J)=1000J (116)(4000J)=150J

none of the above

Part B:

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:

(2)(2000m/s)=4000m/s (2?)(2000m/s)=2800m/s 2000m/s (12?)(2000m/s)=1400m/s (12)(2000m/s)=1000m/s none of the above

Explanation / Answer

Part (A)

Answer is (C) 4000 Joules

Because:

Each and every particle in 1 mole of the gas have diffreent velocity and hence diffrent kinetic energy. The KE(average) of the gas paticles is directly proportional to the absolute Temperature of the gas( in Kelvin scale). The total KE of one mole of gas will be then given by multiplying KE(average) and no of moles.

(KE) = 1/2 mv2 = 3/2 KT

Part(b)

Answer is (f) None of the above.

V(rms) is proportional to the square root of absolute temperature.

Intiall at 50 deC= 323 K

1/2mv2(rms) = 3/2 K (323 K) (1)

so at 1000 deg = 273 K

1/2mv2(rms-new) = 3/2 K (373 K) (2)

eq(2) divided by eq(1) will give:

V(rms-new) = 2000 x sqrt (373/323) = 2149.2 m/s

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