A sample of helium gas at 25 celsius is compressed from 200mL to 0.240 mL Its pressure is now 300mm Hg What was the original pressure of the helium?

Answer 1

Ideal Gas relation

PV=nRT

n1=n2

ASSUMING T1=T2, is this true?

P1*V1=P2*V2

P1=0.24mL*300mmHg/200mL

P1=0.36mmHg

Answer 2

To solve for the original pressure of the helium gas, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature. Using this law, we can set up the equation (P1)(V1) = (P2)(V2), where P1 is the original pressure, V1 is the original volume, P2 is the final pressure, and V2 is the final volume. Plugging in the values gives us (P1)(200 mL) = (300 mm Hg)(0.240 mL). Solving for P1 gives us P1 = (300 mm Hg)(0.240 mL) / 200 mL = 0.36 mm Hg. Therefore, the original pressure of the helium gas was 0.36 mm Hg.