If `10 g` of a gas at atmospheric pressue is cooled from `273^(@)C` to `0^(@)C`, keeping the volume constant, its pressure would become

Options

A. `(1)/(2) "atm"`
B. `(1)/(273) "atm"`
C. 2 atm
D. 273 atm

Answer

To solve the problem of how the pressure of a gas changes when it is cooled from \(273^\circ C\) to \(0^\circ C\) at constant volume, we can use the ideal gas law. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \(P\) = pressure of the gas
- \(V\) = volume of the gas
- \(n\) = number of moles of the gas
- \(R\) = ideal gas constant
- \(T\) = absolute temperature in Kelvin

### Step 2: Convert Temperatures to Kelvin
We need to convert the Celsius temperatures to Kelvin:
- Initial temperature \(T_1 = 273^\circ C = 273 + 273 = 546 \, K\)
- Final temperature \(T_2 = 0^\circ C = 0 + 273 = 273 \, K\)

### Step 3: Use the Relationship of Pressure and Temperature
Since the volume is constant, we can use the relationship between pressure and temperature:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \(P_1\) is the initial pressure (atmospheric pressure, which is approximately \(1 \, atm\))
- \(P_2\) is the final pressure we want to find.

### Step 4: Rearrange the Equation to Solve for \(P_2\)
Rearranging the equation gives us:
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]

### Step 5: Substitute the Values
Substituting the known values into the equation:
\[ P_2 = 1 \, atm \times \frac{273 \, K}{546 \, K} \]
\[ P_2 = 1 \, atm \times \frac{1}{2} = 0.5 \, atm \]

### Final Answer
The final pressure of the gas after cooling from \(273^\circ C\) to \(0^\circ C\) at constant volume is \(0.5 \, atm\).

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