Please answer 5a, 5b and 6 If the volume of a gas is 25.8 mL at 28.7 degree C an

Question

Please answer 5a, 5b and 6

If the volume of a gas is 25.8 mL at 28.7 degree C and 728 mmHg, what is its volume at 15.2 degree C and 757 mmHg? You should be able to do this calculation by reading the experimental write-up. Examine the diagram below. The height of the water column is 18.2 cmH_2O. Convert this to units of mmHg using dimensional analysis. (1 mmHg is equivalent to 13.5 mmH_2O). If the barometric pressure is 762.3 mmHg, what is the pressure of the wet H_2 gas in mmHg? If the pressure of the wet H_2 gas is 580.4 mmHg and the temperature of the water is 21.2 degree C, what is the pressure of the dry gas in mmHg, and in atm? Show your work.

Explanation / Answer

Q. 4 :

Given

Initial volume (V1) of gas 25.8 mL

Intial T (T1) = 28.7 0C = 28.7 0C + 273.15 =301.85 K

Initial pressure (p1) = 728 mmHg A

Final T (T2) = 15.2 0 C + 273.15 = 288.35 K

Final P (P2) = 757 mmHg

Lets use following equation

P1V1T2 = P2V2T1

We have to find final volume of gas.

Lets rearrange above equation so that we get V2 to the left side of equation.

V2 = P1V1T2/P2VT1

Lets plug given values

V2 = [728 x 25.8 x 288.35 / (757 x 301.85) ]

= 23.70 mL

Ans : Final volume = 23.7 mL

Q. 5

a). Height of water column = 18.2 mmH2O

Given:

1 mmHg = 13.5 mmH2O ;

Here we use above relation to convert mmH2O into mmHg

Height in mmHg = 18.2 mmH2O x (1 mmHg / 13.5 mmH2O)

= 1.35 mmHg

So the height in mmHg = 1.35

b).

Given :

Barometric pressure = 762.3 mmHg

And we have to find pressure of wet H2

We know height of the mercury is 760 mm when the pressure is 1 atm

So the height is related to the pressure.

Now we have to find pressure of H2 gas

Pressure of H2 gas = 762.3 mm x (760 mmHg / 760 mm) = 762.3 mmHg

Question 6 :

Given :

Pressure of H2 gas = 580.4 mmHg T = 21.2 0 C Now we have to find pressure of dry gas.

Pressure of dry gas = pressure of wt gas – pressure of water vapor at 21.2 0C

Lets find pressure of water vapor at this T.

Vapor pressure of water at this T = 18.892 mmHg

Lets find now pressure of dry H2 gas.

=( 580.1 – 18.892 ) mmHg

=561.208 mmHg

So the pressure of the wt H2 gas = 561.208 mmHg

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