Two moles of helium are mixed with n moles of hydrogen. If $$\frac{C_p}{C_v} = \frac{3}{2}$$ for the mixture then the value of n is:

We start by recalling the molar heat-capacities of the two gases in the temperature region where only translational and rotational degrees of freedom are active.

For a monatomic gas (such as helium) the degrees of freedom are three, so

$$C_{v,\,\text{He}}=\frac{3}{2}R,\qquad C_{p,\,\text{He}}=C_{v,\,\text{He}}+R=\frac{5}{2}R.$$

For a diatomic gas (such as hydrogen, H2) the active degrees of freedom are five, giving

$$C_{v,\,\text{H}_2}=\frac{5}{2}R,\qquad C_{p,\,\text{H}_2}=C_{v,\,\text{H}_2}+R=\frac{7}{2}R.$$

Two moles of helium are mixed with $$n$$ moles of hydrogen, so the total molar heat capacities of the mixture are obtained by adding the contributions of each component:

$$C_p^{\text{(tot)}}=2\!\left(\frac{5}{2}R\right)+n\!\left(\frac{7}{2}R\right)=5R+\frac{7n}{2}R,$$ $$C_v^{\text{(tot)}}=2\!\left(\frac{3}{2}R\right)+n\!\left(\frac{5}{2}R\right)=3R+\frac{5n}{2}R.$$

The ratio $$\dfrac{C_p}{C_v}$$ for the whole mixture is the ratio of these two totals because the common factor $$R$$ cancels out:

$$\frac{C_p}{C_v}=\frac{5+\dfrac{7}{2}n}{3+\dfrac{5}{2}n}=\frac{5+3.5n}{3+2.5n}.$$

According to the problem statement this ratio equals $$\dfrac{3}{2}$$, so we set up the equation

$$\frac{5+3.5n}{3+2.5n}=\frac{3}{2}.$$

Now we solve algebraically, showing every step:

First cross-multiply:

$$2\left(5+3.5n\right)=3\left(3+2.5n\right).$$

Expanding each side gives

$$10+7n=9+7.5n.$$

Subtract $$9$$ from both sides:

$$1+7n=7.5n.$$

Now subtract $$7n$$ from both sides:

$$1=0.5n.$$

Finally divide by $$0.5$$:

$$n=2.$$

Therefore the mixture must contain two moles of hydrogen.

Hence, the correct answer is Option C.