A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increased by 320.4 L at the same pressure?

T2 = 7.3°C

Explanation:

To solve this problem we will use Charles law equation i. e,

V1/T1 = V2/T2

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2 = 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C = 320.4 L / T2

T2 = 320.4 L * 6.12°C / 269.7 L

T2 = 1960.85 °C. L / 269.7 L

T2 = 7.3°C