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

-be a crystalline solid

- be soluble in water

-conduct electricity

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for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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