Question

4
Calculate the temperature at which
the average velocity of oxygen would
be three times its average at 27 °C. (R =
8.314 JK-1 mol-1, 0 = 16)
1700 K
2700 K
0000
1000 K
3700 K​

Answer 1

Answer:

27

∘

C=300 K

Root mean square speed =

M

3RT

R=8.314×10

7

erg K

−1

mol

−1

;M=32 g mol

−1

;T=300 K

Substituting the values,

=

32

3×8.314×10

7

×300

=483.56 cm/sec

=483.56 m/sec

Average speed =

πM

8RT

=

22×32

8×8.314×10

7

×300×7

=445.42 cm/sec

=445.42 m/sec

Most probable speed =

M

2×R×T

=

32

2×8.314×10

7

×300

=39482 cm/sec

=394.83 m/sec.

Answer 2

The temperature in the final case is 2700K. And option 'B' is correct.

Given:-

Temperature 1 = 27°C

Average velocity at case 2 = 3 × Average velocity at case 1

To Find:-

The temperature in the final case.

Solution:-

We can easily find out the value of temperature in final case by using these simple steps.

As

Temperature 1 = 27°C = 300K

Average velocity at case 2 = 3 × Average velocity at case 1

Temperature 2 =?

According to the formula of Mean velocity of the atoms,

$v(avg) = \sqrt{ \frac{8rt}{\pi \: m} }$

Here in this case, mean velocity in case 1 = case 2

$v(avg)1 = v(avg)2$

$\sqrt{ \frac{8rt2}{\pi \: m} } = 3 \times \sqrt{ \frac{8rt1}{\pi \: m} }$

on comparing,

$\sqrt{t2} = 3 \sqrt{t1}$

on taking squares on both sides,

$t2 = {3}^{2} \times t1$

$t2 = 9 \times t1$

on putting the values,

$t2 = 9 \times 300$

$t2 = 2700k$

Hence, The temperature in the final case is 2700K. And option 'B' is correct.

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