Question Details

The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21x 10⁻²¹ J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)

Options

A

1242x10⁻²¹ J, 968 m/s

B

8.78x10⁻²¹ J, 684 m/s

C

6.21x10⁻²¹ J, 968 m/s

D

12.42x10⁻²¹ J, 684 m/s

Correct Answer :

12.42x10⁻²¹ J, 684 m/s

Solution :

The correct option is 12.42x10⁻²¹ J, 684 m/s.

To find the average translational kinetic energy and the root mean square (rms) speed at the new temperature, we can use the principles of the kinetic theory of gases.

1. Average Translational Kinetic Energy:
The average translational kinetic energy (E) of a gas molecule at an absolute temperature T is given by the formula:
E=32kBT
where kB is the Boltzmann constant. From this formula, we can see that the average translational kinetic energy is directly proportional to the absolute temperature:
ET
Let E1 be the energy at temperature T1=300 K and E2 be the energy at temperature T2=600 K. We can set up the ratio:
E2E1=T2T1
Substituting the given values:
E2=E1×T2T1
E2=(6.21×10-21 J)×600 K300 K
E2=6.21×10-21×2
E2=12.42×10-21 J

2. Root Mean Square (rms) Speed:
The rms speed (vrms) of gas molecules is given by the formula:
vrms=3kBTm
where m is the mass of a single molecule. This shows that the rms speed is directly proportional to the square root of the absolute temperature:
vrmsT
Let v1 be the rms speed at T1=300 K and v2 be the rms speed at T2=600 K. We can express the ratio as:
v2v1=T2T1
Substituting the given values:
v2=v1×T2T1
v2=484 m/s×600 K300 K
v2=484×2 m/s
Using the approximation 21.414:
v2484×1.414 m/s684.38 m/s
Thus, the rms speed at 600 K is approximately 684 m/s.

Combining both results, the values at 600 K are 12.42x10⁻²¹ J and 684 m/s respectively.

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