The Kelvin temperature of a sample of 678 cm3 ammonia gas is doubled, what is the new volume?

To determine the new volume of the sample of ammonia gas when its Kelvin temperature is doubled, we can use Charles's Law. Charles's Law states that, for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin.

Mathematically, Charles's Law is expressed as: \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Where: - \( V_1 \) is the initial volume - \( T_1 \) is the initial temperature - \( V_2 \) is the final volume - \( T_2 \) is the final temperature

Since the problem states that the Kelvin temperature is doubled, we have \( T_2 = 2T_1 \).

The initial volume (\( V_1 \)) is given as 678 cm³, and we want to find the final volume (\( V_2 \)).

Using Charles's Law: \[ \frac{678 \text{ cm}^3}{T_1} = \frac{V_2}{2T_1} \]

To find \( V_2 \), we can rearrange the equation: \[ V_2 = 2 \times 678 \text{ cm}^3 \]

So, the new volume \( V_2 \) will be: \[ V_2 = 1356 \text{ cm}^3 \]

Therefore, when the Kelvin temperature of the sample of ammonia gas is doubled, the new volume of the gas will be 1356 cm³, assuming that the pressure remains constant.