Volume of a given mass of gas at 17 ∘ C is measured as 200 cm 3 . The volume of the same mass of gas at same pressure and a temperature of 47 ∘ C will be ?

Options

A. `77 . 5 cm^(3)`
B. `13 . 45 cm^(3)`
C. `220 . 6 cm^(3)`
D. ` 320 cm^(3)`

Explanation

To solve the problem of finding the volume of a gas at a different temperature while keeping the pressure constant, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is held constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume \( V_1 = 200 \, \text{cm}^3 \)
- Initial temperature \( T_1 = 17^\circ C \)
- Final temperature \( T_2 = 47^\circ C \)

2. **Convert Temperatures to Kelvin:**
- To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
- For \( T_1 \):
\[ T_1 = 17 + 273 = 290 \, K \]
- For \( T_2 \):
\[ T_2 = 47 + 273 = 320 \, K \]

3. **Apply Charles's Law:**
- Charles's Law states:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Rearranging to find \( V_2 \):
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]

4. **Substitute the Values:**
- Substitute \( V_1 \), \( T_1 \), and \( T_2 \) into the equation:
\[ V_2 = 200 \, \text{cm}^3 \times \frac{320 \, K}{290 \, K} \]

5. **Calculate \( V_2 \):**
- Perform the calculation:
\[ V_2 = 200 \times \frac{320}{290} \approx 220.69 \, \text{cm}^3 \]

6. **Round the Answer:**
- Rounding to one decimal place gives:
\[ V_2 \approx 220.7 \, \text{cm}^3 \]

### Final Answer:
The volume of the same mass of gas at a temperature of 47 °C will be approximately **220.7 cm³**.

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