Question

1.5 mole of monoatomic gas is mixed with 3.5 moles of diatomic gas .The molecular specific heat of mixture at constant V is x×10^-2 mole k-1. Then x is​

Answer 1

Answer: 4.5

Explanation:

As we know, Cv for monoatomic gas is 3R/2 and for diatomic gas is 5R/2.

So for mixture, Cv=41×23R+23×5R=49R=49×2=4.5 cal

Answer 2

Answer:

The value of x is 1829.08.

Explanation:

The molecular specific heat of the mixture of gas at constant volume is given as,

$C_v_m_i_x=\frac{C_v_1 n_1+C_v_2n_2}{n_1+n_2}$      (1)

Where,

Cvmix= molecular specific heat of mixture at constant volume

Cv₁=molecular specific heat of the first gas

n₁=number of moles of the first gas

Cv₂=molecular specific heat of the first gas

n₂=number of moles of the second gas

From the question we have,

n₁=1.5

n₂=3.5

Cv₁=3R/2   (as the gas is monoatomic)

Cv₂=5R/2  (as the gas is diatomic)

R=universal gas constant=8.314kpa.dm³mol⁻¹K⁻¹

Cvmix=x×10⁻²mole k⁻¹         (2)

By substituting n₁,n₂, Cv₁, and Cv₂ into equation (1);

$C_v_m_i_x=\frac{\frac{3R}{2}\times 1.5+\frac{5R}{2}\times 3.5}{1.5+3.5}=\frac{11R}{5}$     (3)

By equating equations (2) and (3) we get;

$x\times 10^-^2=\frac{11R}{5}$

$x=\frac{1100R}{5}=220R=1829.08$

Hence, the value of x is 1829.08.