A container contains 1 mole of a gas at 1 atm pressure and 27° C, while its volume is 24.6 litres. If its pressure is 10 atm and temperature 327° C, then new volume is

Options

A. 2.56 litres
B. 3.15 litres
C. 4.92 litres
D. 5.44 litres

Answer

To find the new volume of the gas under the new conditions, we can use the Ideal Gas Law and the combined gas law. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Initial pressure, \( P_1 = 1 \, \text{atm} \)
- Initial volume, \( V_1 = 24.6 \, \text{L} \)
- Initial temperature, \( T_1 = 27^\circ C \)
- Final pressure, \( P_2 = 10 \, \text{atm} \)
- Final temperature, \( T_2 = 327^\circ C \)

### Step 2: Convert temperatures from Celsius to Kelvin
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 \):
\[ T_1 = 27 + 273 = 300 \, K \]
- For \( T_2 \):
\[ T_2 = 327 + 273 = 600 \, K \]

### Step 3: Use the combined gas law
The combined gas law is given by:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
We can rearrange this to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 4: Substitute the known values into the equation
Now, substitute the known values into the equation:
\[ V_2 = \frac{(1 \, \text{atm}) \times (24.6 \, \text{L}) \times (600 \, K)}{(10 \, \text{atm}) \times (300 \, K)} \]

### Step 5: Calculate \( V_2 \)
Calculating the numerator:
\[ 1 \times 24.6 \times 600 = 14760 \]
Calculating the denominator:
\[ 10 \times 300 = 3000 \]
Now, divide the numerator by the denominator:
\[ V_2 = \frac{14760}{3000} = 4.92 \, \text{L} \]

### Conclusion
The new volume \( V_2 \) of the gas is **4.92 liters**.

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