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 relationship between pressure and temperature given by Gay-Lussac's Law. Here's a step-by-step breakdown of the solution:

### Step 1: Convert Temperatures to Kelvin
The temperatures need to be converted from Celsius to Kelvin since the gas laws require absolute temperature.

- **T1 (Initial Temperature)**:
\[ T1 = 273°C = 273 + 273 = 546 \text{ K} \]

- **T2 (Final Temperature)**:
\[ T2 = 0°C = 0 + 273 = 273 \text{ K} \]

### Step 2: Identify Initial Pressure
The initial pressure (P1) is given as 1 atmosphere.

- **P1 = 1 atm**

### Step 3: Apply Gay-Lussac's Law
According to Gay-Lussac's Law, the pressure of a gas at constant volume is directly proportional to its absolute temperature. This can be expressed mathematically as:
\[ \frac{P1}{T1} = \frac{P2}{T2} \]

### Step 4: Rearrange the Equation to Solve for P2
We can rearrange the equation to find the final pressure (P2):
\[ P2 = P1 \times \frac{T2}{T1} \]

### Step 5: Substitute the Known Values
Now we can substitute the known values into the equation:
\[ P2 = 1 \text{ atm} \times \frac{273 \text{ K}}{546 \text{ K}} \]

### Step 6: Calculate P2
Now we perform the calculation:
\[ P2 = 1 \text{ atm} \times \frac{273}{546} = 1 \text{ atm} \times \frac{1}{2} = 0.5 \text{ atm} \]

### Final Answer
The pressure of the gas after cooling from 273°C to 0°C at constant volume is:
\[ \boxed{0.5 \text{ atm}} \]