Question

On a day when the barometer reads 75.23 cm, a reaction vessel holds 250 mL of ideal gas at 20 celsius. An oil manometer ( rho= 810 kg/m^3) reads the pressure in the vessel to be 41 cm of oil and below atmospheric pressure. What volume will the gas occupy under S.T.P.?

Answer 1

Answer:

V2 = 8.25 ml

Explanation:

First we will list down the data that is given to us:

V1 = 250 ml

T1 = 20° C +273K= 293K

T2= 25° C + 273K = 298K ( AT STP)

Density at STP = 13600 kg/m^3

First we will calculate the pressure inside the vessel using the formula

P = (ρ)(g)(h)

P1 = (810)(9.8)(0.41)

P1 = 3254.58 pa

P2 = (13600)(9.8)(0.7523)

P2 = 100266.544 pa

Now in order calculate the volume that the gas occupies we will use the following formula,

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

Making V2 as the subject of the equation we get,

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

V2 = (3254.58*250*298)/(100266.544*293)

V2 = 8.25 ml