Question

3 moles of O2 and 5 most of AR are present in a closed container find sum of the internal energy in terms of RT
15RT
10RT
5RT
20RT ​

Answer 1

Answer:

$\boxed{\mathfrak{ Internal \ energy \ (U) = 15RT}}$

Given:

Moles of Ar ($\sf n_{Ar}$) = 5

Moles of $\sf O_2$ ($\sf n_{O_2}$) = 3

To Find:

Initial energy of the system (U)

Explanation:

Degree of freedom of Ar ($\sf f_{Ar}$) = 3

Degree of freedom of $\sf O_2$ ($\sf f_{O_2}$) = 5

Internal energy of gaseous molecule:

$\boxed{ \bold{U = \frac{f}{2}nRT }}$

Where,

f $\rightarrow$ Degree of freedom

n $\rightarrow$ Moles of gas

R $\rightarrow$ Gas constant

T $\rightarrow$ Temperature

So,

Internal energy of the given system:

$\sf \implies U = U_{Ar} + U_{O_2} \\ \\ \sf \implies U = \frac{f_{Ar} }{2}n_{Ar} RT + \frac{f_{O_2} }{2}n_{O_2} RT \\ \\ \sf \implies U = \frac{3}{2} \times 5RT + \frac{5}{2} \times 3RT \\ \\ \sf \implies U = \frac{15}{2} RT + \frac{15}{2}RT \\ \\ \sf \implies U = \frac{15 + 15}{2} RT \\ \\ \sf \implies U = \frac{30}{2} RT \\ \\ \sf \implies U = 15 RT$

Answer 2

Answer:

U = 5/2 × 3 RT + 3/2 × 5 RT

U = 7.5 RT + 7.5 RT

U = 15 RT