Calculate the pressure of gas at constant volume , if a 10g of a gas at one atmospheric pressure is cooled from `273^@C " to "0^@C`

Options

A. 1/2 atm
B. 1/273 atm
C. 2 atm
D. 273 atm

Answer

To solve the problem of calculating the pressure of a gas at constant volume when it is cooled from 273°C to 0°C, we can use the ideal gas law, specifically the relationship between pressure and temperature at constant volume.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The initial temperature \( T_1 \) is given as \( 273^\circ C \).
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- Thus,
\[ T_1 = 273 + 273.15 = 546.15 \, K \]
- The final temperature \( T_2 \) is \( 0^\circ C \):
\[ T_2 = 0 + 273.15 = 273.15 \, K \]

2. **Identify Initial Pressure:**
- The initial pressure \( P_1 \) is given as \( 1 \, atm \).

3. **Use the Formula for Pressure and Temperature Relationship:**
- At constant volume, the relationship between the initial and final pressures and temperatures can be expressed as:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
- Rearranging this formula to find \( P_2 \) gives:
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]

4. **Substitute the Known Values:**
- Substitute \( P_1 = 1 \, atm \), \( T_1 = 546.15 \, K \), and \( T_2 = 273.15 \, K \):
\[ P_2 = 1 \, atm \times \frac{273.15 \, K}{546.15 \, K} \]

5. **Calculate \( P_2 \):**
- Performing the calculation:
\[ P_2 = 1 \times \frac{273.15}{546.15} \approx 0.5 \, atm \]

### Final Answer:
The pressure of the gas after cooling to \( 0^\circ C \) is approximately \( 0.5 \, atm \).