Chemistry, 19.04.2020 00:48 st23pgardner

A mixture of 6.00 g of O2 (g) and 9.00 g of CH4 (g) is placed in a 15.0 L vessel at 0o C. What is the partial pressure of each gas? What is the total pressure in the vessel?

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Answer from: Ully4559

Answer : The partial pressure of $O_2\text{ and }CH_4$ is, 0.281 atm and 0.839 atm respectively.

The total pressure in vessel is, 1.12 atm.

Explanation :

First we have to calculate the moles of oxygen and methane.

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}=\frac{6.00g}{32g/mol}=0.188mol$

and,

$\text{Moles of }CH_4=\frac{\text{Given mass }CH_4}{\text{Molar mass }CH_4}=\frac{9.00g}{16g/mol}=0.562mol$

Now we have to calculate the total moles of gas.

Total moles of gas = Moles of oxygen + Moles of methane

Total moles of gas = 0.188 + 0.562

Total moles of gas = 0.75 mol

Now we have to calculate the total pressure of gas by using ideal gas equation.

PV=nRT

where,

P = total pressure of gas = ?

V = total volume of gas = 15.0 L

n = total number of moles gas = 0.75 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 0^oC=273+0=273K

Putting values in above equation, we get:

$P\times 15.0L=0.75mole\times (0.0821L.atm/mol.K)\times 273K$

P=1.12atm

Now we have to calculate the mole fraction of oxygen and methane.

$\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }O_2+\text{Moles of }CH_4}=\frac{0.188}{0.188+0.562}=0.251$

and,

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }O_2+\text{Moles of }CH_4}=\frac{0.562}{0.188+0.562}=0.749$

Now we have to calculate the partial pressure of oxygen and methane.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

p_i = partial pressure of gas

p_T = total pressure of gas = 1.12 atm

X_i = mole fraction of gas

$p_{O_2}=X_{O_2}\times p_T$

$p_{O_2}=0.251\times 1.12atm=0.281atm$

and,

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.749\times 1.12atm=0.839atm$

Thus, the partial pressure of $O_2\text{ and }CH_4$ is, 0.281 atm and 0.839 atm respectively and the total pressure in vessel is, 1.12 atm.

Answer from: Quest

a mercury barometer is a device that uses the physical properties of mercury in a tube to measure barometric pressure. mercury barometers are used for : -

1. measurement of atmospheric pressure

2. determining the altitude of a place

3. weather forecasting

4. calibration and checking of aneroid barometers

5. application in fluid mechanics

6. measurement of pressure in aircrafts

7. surface weather analysis

8. preparation of barographs

9. usage in physical sciences like physics, astronomy & chemistry

10. preparation of aircraft altimeters

characteristics of mercury barometer : -

1.used to define and measure atmospheric pressure.

2.on the average at sea level the column of mercury rises to a height of about 760 mm.

3.this quantity is equal to 1 atmosphere.

4.it is also known as standard atmospheric pressure.

5.a barometer in which the weight of a column of mercury in a glass tube with a sealed top is balanced against that of the atmosphere pressing on an exposed cistern of mercury at the base of the mercury column, the height of the column varying with atmospheric pressure.

Write 15 uses of a mercury barometer along with its characterestics

Answer from: Quest

100%sorry if i’m wronghope this good luck: )

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A mixture of 6.00 g of O2 (g) and 9.00 g of CH4 (g) is placed in a 15.0 L vessel at 0o C. What is th...

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