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5.1g of NH3 and 3g of NO are mixed and kept in a container of 1L at 27°c.find total pressure exerted by gases

Answers

Answered by Sumitnegi58

Answer:

Explanation:

Explanation:In other words, the total pressure is the sum of the individual partial pressures.And Dalton's Law of partial pressures

Law of partial pressures assures us that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container.And we can use the Ideal Gas equation to estimate the pressure:

VSo all we have to do is to solve for P gas

gas individually:

gas individually:P

2=1.00

2=1.00⋅

2=1.00⋅g

2=1.00⋅g2.016

2=1.00⋅g2.016⋅

2=1.00⋅g2.016⋅g

2=1.00⋅g2.016⋅g⋅

2=1.00⋅g2.016⋅g⋅m

2=1.00⋅g2.016⋅g⋅mo

2=1.00⋅g2.016⋅g⋅mol

2=1.00⋅g2.016⋅g⋅mol−

2=1.00⋅g2.016⋅g⋅mol−1

2=1.00⋅g2.016⋅g⋅mol−1×

2=1.00⋅g2.016⋅g⋅mol−1×0.0821

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅a

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅at

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atm

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅m

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mo

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅a

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅at

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅atm

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅atmP

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅atmPH

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅atmPHe

2=1.00⋅g2.016⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mol×300⋅K1.00⋅L≅12⋅atmPHe=1.00⋅g4.00⋅g⋅mol−1×0.0821⋅L⋅atmK⋅mo

tmK⋅mol×300⋅K1.00⋅L

300⋅K1.00⋅L≅6⋅atm.

300⋅K1.00⋅L≅6⋅atm.They are spoon feeding you a bit here, in that they explicitly tell you that hydrogen is a bimolecular gas; in fact all the elemental gases SAVE THE NOBLE GASES are bimolecular:

Alternatively, I could have summed the total moles of gas, and plugged that quantity into the Ideal Gas Equation. The partial pressure of each gas would be proportional to the mole fraction.

Answered by Anonymous

Given:

No: of moles of Ammonia( $NH_3$ ) = 5.1 g
No: of moles of Nitrogen Oxide(NO) = 3 g
Volume(V) = 1 L
Temperature(T) = 27°C = 27° C + 273 K = 300 K

To Find:

The total pressure exerted by the gas(P).

Solution:

There are two different gases given. To find the total pressure exerted by them, we should find the pressure of both the gases individually.
Finding the pressure of Ammonia( $NH_3$ ) by using Ideal gas equation, $P_1 =\frac{nRT}{V}$
Where, n is the no: of moles of gas, R is the molar gas Constant(R = 0.083 $bar dm^{3}K^{-1}mol^{-1}$ ), T is the temperature and V is the volume.
Substituting the values we get, $P_1 = \frac{5.1*0.083*300}{1} = 126.99 bar$
$P_1$ = 126.99 bar
finding the pressure of Nitrogen Oxide(NO), $P_2 = \frac{nRT}{V}$
Substituting the values we get, $P_2 = \frac{3*0.083*300}{1} = 74.7 bar$
$P_2$ = 74.7 bar.
Total pressure of the gas P = $P_1+P_2$ = 126.99 + 74.7
Total Pressure, P = 201.69 bar.

Total pressure exerted by gas, P = 201.69 bar.

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5.1g of NH3 and 3g of NO are mixed and kept in a container of 1L at 27°c.find total pressure exerted by gases ​

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5.1g of NH3 and 3g of NO are mixed and kept in a container of 1L at 27°c.find total pressure exerted by gases