Find the temperature at which root mean square velocity of a gas is half of its value at zero degree celsius
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See, T is directely proportional to c^2. So,
Let the initial temparture is T1 = 0 C OR 273K And the root mean square velocity at this temparture = c (SMALL C)
And final temperature be T2 =? And the root mean square velocity =C =c/2 (given in question)
T1/T2 = c^2/C^2
=T2 = T1 *C^2/ c^2
=T2 = 273 * (c/2)^2 /c^2
=T2 = 273 * c^2/4c^2
=T2 =273 /4 = 68.25 K
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Let the initial temparture is T1 = 0 C OR 273K And the root mean square velocity at this temparture = c (SMALL C)
And final temperature be T2 =? And the root mean square velocity =C =c/2 (given in question)
T1/T2 = c^2/C^2
=T2 = T1 *C^2/ c^2
=T2 = 273 * (c/2)^2 /c^2
=T2 = 273 * c^2/4c^2
=T2 =273 /4 = 68.25 K
HOPE IT HELPS
Please mark this answer as a brainlist answer if you like the answer please
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