Suppose you have 4.0 L of helium in a flexible container (like a balloon) at a certain temperature and pressure. What will the volume of the helium be if the absolute temperature is halved and the pressure is doubled?

When the temperature is halved and the pressure is doubled, the volume will be divided by 4. The new volume will be 1.0 L

Explanation:

Step 1: Data given

Volume of the balloon = 4.0 L

Suppose the pressure of helium gas = 100 atm

Suppose the temperature = 100 K

The absolute temperature is halved = 50 K

The pressure is doubled = 200 atm

Step 2: Calculate the new volume

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 100 atm

⇒with V1 = the initial volume = 4.0 L

⇒with T1 = the initial temperature = 100 K

⇒with P2 = the doubled pressure = 200 atm

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the halved temperautre = 50 K

(100 atm * 4.0 L) / 100 K = (200 atm * V2) / 50 K

V2 = (100 * 4 * 50 ) / (100*200)

V2 = 1 L

When the temperature is halved and the pressure is doubled, the volume will be divided by 4. The new volume will be 1.0 L

it is stable as it is in group 0 which have full outer shells of electrons