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2 days ago

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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= 8

10 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.?

1 answer:

kicyunya [1K]2 days ago

8 0

Answer:

V2 = 8.25 ml

Explanation:

Let’s first outline the provided data:

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 find the pressure within 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 to determine the volume occupied by the gas, we apply the formula:

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

To solve for V2, we rearrange it as follows:

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

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

V2 = 8.25 ml

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