The volume of a gas at `20^(@)C` is 200 ml. If the temperature is reduced to `-20^(@)C` at constant pressure, its volume will be :-

Options

A. (a)172.6 ml
B. (b)17.26 ml
C. (c)192.7 ml
D. (d)19.27 ml

Answer

To solve the problem, we will use the Charles' Law, which states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin). The formula can be expressed as:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 1: Convert temperatures from Celsius to Kelvin
- The initial temperature \( T_1 \) is \( 20^\circ C \).
- To convert to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_1 = 20 + 273 = 293 \, K \]

- The final temperature \( T_2 \) is \( -20^\circ C \).
Using the same conversion:
\[ T_2 = -20 + 273 = 253 \, K \]

### Step 2: Write down the known values
- Volume \( V_1 = 200 \, ml \)
- Temperature \( T_1 = 293 \, K \)
- Temperature \( T_2 = 253 \, K \)
- We need to find \( V_2 \).

### Step 3: Apply Charles' Law
Using the formula:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Substituting the known values:
\[ \frac{200 \, ml}{293 \, K} = \frac{V_2}{253 \, K} \]

### Step 4: Solve for \( V_2 \)
Cross-multiplying gives:
\[ V_2 = \frac{200 \, ml \times 253 \, K}{293 \, K} \]

### Step 5: Calculate \( V_2 \)
Now we perform the calculation:
\[ V_2 = \frac{50600 \, ml \cdot K}{293 \, K} \approx 172.6 \, ml \]

### Final Answer
Thus, the volume of the gas when the temperature is reduced to \(-20^\circ C\) at constant pressure is approximately \( 172.6 \, ml \).

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