Ahelium balloon has a volume of 500ml at stp. what must the temperature be, in °c if the volume increased to 750 ml and the pressure decreased to .5 atm

The temperature must be -49.5°C

Explanation:

A helium balloon has a volume of 500 mL (=0.5L) at stp this means 25°C and 1atm

In this situation, the volume is increased to 750 mL (= 0.750L) and the pressure decreased to 0.5 atm

For this situation we will use the following formula:

(P1 * V1)/T1  = (P2 * V2) /T2

with P1 = initial pressure = 1atm

with V1 = the initial volume = 0.5 L

with T1 = the initial temperature = 25°C = 298.15 Kelvin

with P2= The changed pressure = 0.5 atm

with V2 = The changed volume = 0.750 L

with T2 = The changed temperature = TO BE DETERMINED

(P1 * V1)/T1  = (P2 * V2) /T2

(1atm *0.5L)/298.15 K  = (0.5 atm * 0.750 L) / T2

T2 = 223.6 Kelvin = -49.5 °C

The temperature must be -49.5°C

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