The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. What are the pressures of the gases for equal number of moles?

For real gases we use the relation $$PV = Z\,nRT$$, where $$P$$ is pressure, $$V$$ is volume, $$n$$ is the number of moles, $$T$$ is the temperature, $$R$$ is the universal gas constant and $$Z$$ is the compressibility factor.

Both gases A and B are taken at the same temperature and contain an equal number of moles, so $$n$$, $$R$$ and $$T$$ are common to both. Hence we can write

$$P_A V_A = Z_A nRT \quad\text{and}\quad P_B V_B = Z_B nRT.$$

Dividing the first equation by the second we get

$$\dfrac{P_A V_A}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

The problem tells us that the volume of gas A is twice that of gas B, so $$V_A = 2V_B.$$ Substituting this into the ratio gives

$$\dfrac{P_A (2V_B)}{P_B V_B} = \dfrac{Z_A}{Z_B}.$$

Simplifying the left side, $$V_B$$ cancels:

$$\dfrac{2P_A}{P_B} = \dfrac{Z_A}{Z_B}.$$

We are also told that the compressibility factor of gas A is three times that of gas B, so $$Z_A = 3Z_B.$$ Substituting this value yields

$$\dfrac{2P_A}{P_B} = \dfrac{3Z_B}{Z_B} = 3.$$

Now solving for the ratio of the pressures, we have

$$2P_A = 3P_B.$$

This is exactly the relation asked for between the two pressures. Hence, the correct answer is Option A.