Question

the average kinetic energy per molecule of an ideal gas at 27°C is —(A)150R,(B)(150R)/NA,(C)450R,(D)(450R)/NA​

Answer 1

Answer:

Factorisation of Polynomials

EXERCISE 3F

Factorise

1. r +27

3. 125a +

1

+ 1

5. 16x* +54x

7. x + x²

9. 1 - 27a

11. x°-512

2. 27a? +64

4.216x? +

125

6. 7a56b3

8. q +0.008

10.64a? - 343

12. a? -0.064

1

X

3

13. 8r

14.

216-8y

3

27v

15. x - 8xy

17. 3a%b-81a464

19. 8x’y'-x

21. x - 729

23. (a+b): - (a - b)

16.

Answer 2

Answer:

(D)(450R)/NA​ is the answer

Explanation:

We know that,

$KE = \dfrac{3}{2} nRT \ \ \ \ \ \text{[For n number of moles]}\\\\So, \ KE = \dfrac{\bigg(\dfrac{3}{2}RT \bigg)}{N_A} \ \ \ \ \ \text{[Per molecule]}$

Here, T is given as 27°C. But in T must be in Kelvin K. So,

Temperature = 27 + 273 = 300K

Thus,

$KE = \dfrac{\dfrac{3}{2} R(300)}{N_A} = \dfrac{450R}{N_A}$

Hope it helps!