Two separate bulbs contains ideal gases P and q, respectively maintained at the same temperature. The density of gas P is twice of that of the Q, and the molecular weight of the gas P is half of that of the gas Q. The ratio of the pressure of gas P to that of gas Q is

Options

A. 2
B. `1//2`
C. 4
D. `1//4`

Explanation

To solve the problem step by step, we will use the ideal gas law and the relationships between density, molecular weight, and pressure.

### Step 1: Understand the given information
- Let the density of gas Q be \( \rho_Q \).
- The density of gas P is twice that of gas Q:
\[ \rho_P = 2\rho_Q \]
- Let the molecular weight of gas Q be \( M_Q \).
- The molecular weight of gas P is half that of gas Q:
\[ M_P = \frac{1}{2} M_Q \]

### Step 2: Use the ideal gas law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature

From the number of moles \( n \), we have:
\[ n = \frac{m}{M} \]
Where \( m \) is the mass of the gas and \( M \) is the molecular weight.

### Step 3: Express pressure in terms of density and molecular weight
Using the relationship of density:
\[ \rho = \frac{m}{V} \]
We can rearrange this to express mass \( m \):
\[ m = \rho V \]
Substituting this into the equation for \( n \):
\[ n = \frac{\rho V}{M} \]
Substituting \( n \) back into the ideal gas law:
\[ PV = \frac{\rho V}{M} RT \]
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ P = \frac{\rho RT}{M} \]

### Step 4: Write the pressure equations for both gases
For gas P:
\[ P_P = \frac{\rho_P RT}{M_P} \]
For gas Q:
\[ P_Q = \frac{\rho_Q RT}{M_Q} \]

### Step 5: Substitute the known relationships
Substituting \( \rho_P = 2\rho_Q \) and \( M_P = \frac{1}{2} M_Q \) into the pressure equations:
\[ P_P = \frac{2\rho_Q RT}{\frac{1}{2} M_Q} = \frac{2 \cdot 2\rho_Q RT}{M_Q} = \frac{4\rho_Q RT}{M_Q} \]
\[ P_Q = \frac{\rho_Q RT}{M_Q} \]

### Step 6: Find the ratio of pressures
Now, we can find the ratio of the pressures:
\[ \frac{P_P}{P_Q} = \frac{\frac{4\rho_Q RT}{M_Q}}{\frac{\rho_Q RT}{M_Q}} = \frac{4\rho_Q RT}{\rho_Q RT} = 4 \]

### Final Result
Thus, the ratio of the pressure of gas P to that of gas Q is:
\[ \frac{P_P}{P_Q} = 4:1 \]

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