One litre of a gas weighs 2 g at 300 K and 1 atm pressure. If the pressure is made 0.75 atm, at which of the following temperature will one litre of the same gas weigh 1 gram : 1) 450K 2) 600K 3) 800K 4)900K

Options

A. 450 K
B. 600 K
C. 800 K
D. 900 K

Explanation

To solve the problem, we can use the Ideal Gas Law and the relationship between pressure, volume, temperature, and the number of moles of a gas. The Ideal Gas Law can be expressed as:

\[ PV = nRT \]

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

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Initial mass of gas (\( W_1 \)) = 2 g
- Initial temperature (\( T_1 \)) = 300 K
- Initial pressure (\( P_1 \)) = 1 atm

2. **Identify the final conditions**:
- Final mass of gas (\( W_2 \)) = 1 g
- Final pressure (\( P_2 \)) = 0.75 atm
- Final temperature (\( T_2 \)) = ? (to be determined)

3. **Calculate the number of moles for initial and final conditions**:
- The number of moles can be expressed as:
\[ n_1 = \frac{W_1}{M} = \frac{2 \text{ g}}{M} \]
\[ n_2 = \frac{W_2}{M} = \frac{1 \text{ g}}{M} \]
- Here, \( M \) is the molar mass of the gas.

4. **Apply the Ideal Gas Law**:
- For the initial state:
\[ P_1 \cdot V = n_1 \cdot R \cdot T_1 \]
\[ 1 \cdot 1 = \frac{2}{M} \cdot R \cdot 300 \]
- For the final state:
\[ P_2 \cdot V = n_2 \cdot R \cdot T_2 \]
\[ 0.75 \cdot 1 = \frac{1}{M} \cdot R \cdot T_2 \]

5. **Set up the equation**:
- From the initial state:
\[ 1 = \frac{2 \cdot R \cdot 300}{M} \]
- From the final state:
\[ 0.75 = \frac{1 \cdot R \cdot T_2}{M} \]

6. **Relate the two equations**:
- Rearranging the first equation gives:
\[ M = \frac{2 \cdot R \cdot 300}{1} \]
- Substitute \( M \) into the second equation:
\[ 0.75 = \frac{1 \cdot R \cdot T_2}{\frac{2 \cdot R \cdot 300}{1}} \]
- Simplifying gives:
\[ 0.75 = \frac{T_2}{600} \]

7. **Solve for \( T_2 \)**:
- Multiply both sides by 600:
\[ T_2 = 0.75 \cdot 600 = 450 \text{ K} \]

### Conclusion:
The temperature at which one litre of the gas weighs 1 gram at a pressure of 0.75 atm is **450 K**. Therefore, the correct answer is option 1.