1 mole of gas occupies a volume of 100 ml at 50 mm pressure . What is the volume occupied by two moles of gas at 100 mm pressure and at same temperature

Options

A. 500 ml
B. 200 ml
C. 100 ml
D. 50 ml

Explanation

To solve the problem, we will use the ideal gas law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature

Given:
- For the first case (1 mole of gas):
- \( P_1 = 50 \, \text{mm} \)
- \( V_1 = 100 \, \text{ml} \)
- \( n_1 = 1 \, \text{mole} \)

- For the second case (2 moles of gas):
- \( P_2 = 100 \, \text{mm} \)
- \( n_2 = 2 \, \text{moles} \)
- \( V_2 = ? \) (This is what we need to find)

Since the temperature is the same in both cases, we can set up the equation using the ideal gas law for both cases:

1. For the first case:
\[ P_1 V_1 = n_1 R T \]

2. For the second case:
\[ P_2 V_2 = n_2 R T \]

Since \( R \) and \( T \) are constant, we can relate the two equations:

\[ \frac{P_1 V_1}{n_1} = \frac{P_2 V_2}{n_2} \]

Substituting the known values:

\[ \frac{50 \, \text{mm} \times 100 \, \text{ml}}{1 \, \text{mole}} = \frac{100 \, \text{mm} \times V_2}{2 \, \text{moles}} \]

Now, we can simplify this equation:

\[ 50 \times 100 = \frac{100 \times V_2}{2} \]

\[ 5000 = 50 \times V_2 \]

Now, divide both sides by 50:

\[ V_2 = \frac{5000}{50} = 100 \, \text{ml} \]

Thus, the volume occupied by 2 moles of gas at 100 mm pressure and at the same temperature is:

\[ \boxed{100 \, \text{ml}} \]