Question

At constant temperature, a sample of helium at 760. torr in a closed container was compressed from 5.00 L to 3.00 L, with no change in amount of gas or temperature. What was the new pressure exerted by the helium on its container

Answer 1

The new pressure is 1.67 atm or 1266.7 torr

Step-by-step explanation:

Step 1: Data given

Pressure in the container = 760 torr = 1 atm

Initial volume = 5.00 L

Final volume = 3.00 L

The amount of gas does not change

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1 atm

⇒with V1 = the initial volume = 5.00 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the new volume = 3.00 L

1 atm * 5.00 L = P2 * 3.00L

P2 = 5.00/3.00

P2 = 1.67 atm

The new pressure is 1.67 atm or 1266.7 torr

Answer 2

The new pressure is 1,67atm

Step-by-step explanation:

In this case, having constant temperature we apply Boyle Mariotte's Law, where the volume varies inversely proportional to the pressure: P1 xV1 = P2xV2. We pass the pressure unit from Torr to atm: 760 Torr = 1atm

P1 xV1 = P2xV2 --> P2=P1 xV1/V2

P2= 1 atm x 5,00L/3, 00L=1, 67 atm