Question

An ideal gas at \( 27^\circ C \) is heated at constant pressure till the volume becomes three times. The temperature of the gas will be

• A. \( 81^\circ C \)
• B. \( 627^\circ C \)
• C. \( 900^\circ C \)
• D. \( 1173^\circ C \)

Hint:

Always remember to use absolute temperatures (in Kelvin) when applying gas laws like Charles's Law. Convert Celsius to Kelvin by adding 273.15 (or approximately 273).

Answer 1

The Correct Option is B

Step 1: Identify the given parameters and the process.
We are given:
Initial temperature \( T_1 = 27^\circ C \)
The process is at constant pressure (\( P_1 = P_2 \))
Final volume \( V_2 = 3 V_1 \), where \( V_1 \) is the initial volume. We need to find the final temperature \( T_2 \).
Step 2: Apply Charles's Law for a constant pressure process.
For an ideal gas undergoing a constant pressure process, Charles's Law states that the volume is directly proportional to the absolute temperature:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
where \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin.
Step 3: Convert the initial temperature to Kelvin.
To use Charles's Law, we need to convert the temperature from Celsius to Kelvin:
$$T_1 (\text{K}) = T_1 (^\circ C) + 273.15$$
$$T_1 = 27 + 273.15 = 300.15 \text{ K}$$
For simplicity in calculations with the given options, we can often use \( 273 \) instead of \( 273.15 \). So, \( T_1 \approx 27 + 273 = 300 \text{ K} \). Step 4: Solve for the final temperature \( T_2 \).
Using Charles's Law:
$$\frac{V_1}{300 \text{ K}} = \frac{3 V_1}{T_2}$$We can cancel \( V_1 \)
from both sides:$$\frac{1}{300} = \frac{3}{T_2}$$
$$T_2 = 3 \times 300 \text{ K} = 900 \text{ K}$$
Step 5: Convert the final temperature back to Celsius.
$$T_2 (^\circ C) = T_2 (\text{K}) - 273.15$$
$$T_2 \approx 900 - 273 = 627^\circ C$$
Therefore, the final temperature of the gas will be approximately \( 627^\circ C \).