Question

If 2 mole of an ideal monatomic gas at temperature T₀ is mixed with 4 moles of another ideal monatomic gas at temperature 2T₀, then the temperature of the mixture is
(a) $\frac{5}{3}T_{0}$
(b) $\frac{3}{2}T_{0}$
(c) $\frac{4}{3}T_{0}$
(d) $\frac{5}{4}T_{0}$

Answer 1

Answer:

OPTION A

Explanation:

LOOK AT SOLUTION ATTACHED

Answer 2

The temperature of the mixture is

(a)  5T₀/3

Let the final temperature of the mixture is T'.

So, the total energy is conserved after mixing.

For the first gas,

Internal energy U₁ = (f₁/2) × n₁ × R × T₁

f is the degrees of freedom,

R is universal gas constant,

T is the temperature

n is no. of moles of gas.

Similary for second gas,

U₂ = (f₂/2) × n₂ × R × T₂

After mixing, final energy of the first gas is,

U₁' = (f₁/2) × n₁ × R × T

Similarly, after mixing, final energy of second gas is,

U₂' = (f₂/2) × n₂ × R × T

So,

U₁ + U₂ = U₁' + U₂'

⇒[(f₁/2) × n₁ × R × T₁] + [(f₂/2) × n₂ × R × T₂] = [(f₁/2) × n₁ × R × T] + [(f₂/2) × n₂ × R × T]

As both the gases are monoatomic, so f₁ = f₂ and R are constant, so they are cancelled from both sides of the equation.

⇒  n₁T₁ + n₂T₂ = (n₁ + n₂)T

n₁ = 2 moles,

n₂ = 4 moles

T₁ = T₀

T₂ = 2T₀

Putting their values, we get:

2T₀ + 8T₀ = 6T

⇒ T = 10T₀/6 =  5T₀/3

Option(A) is correct.