Question

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?

Answer 1

Answer:

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