Question

At what temperature the average speed of gas molecules will be double that at 27 degree celsius

Answer 1

Answer:

Average speed,

V=√(8R*T/πM).

At 27 C.

V1= √(8R*27/πM).

Let V2 be the speed which is double of V1 so,

V1/V2=√(8R*T/πM)/√(8R*27/πM).

1/2=√27/√T2.

Squaring on both the sides we get,

T2=27*4.

T2=108 C.

Step-by-step explanation:

Answer 2

Given - Initial temperature

Find - Final temperature

Solution - At 108° C the average speed of gas molecules will double.

The speed of gas molecules and temperature are related to each other as -

v = ✓T

Now, forming relation for initial and final temperature.

V1/V2 = ✓T1/T2

As per the question, V2 = 2V1

Now, keep the values in the formula to find the value of T2.

V1/2V1 = ✓27/T2

1/2 = ✓27/T2

27/T2 = (1/2)²

27/T2 = 1/4

T2 = 27*4

T2 = 108

So, at 108° C, the average speed of gas molecules will double.