Chemistry, asked by vipinmeena53, 11 months ago

in a vessel 3 mole and 2 mole 2 gases A and B are present if partial pressure of A is 5atm then calculate partial pressure of B​

Answers

Answered by adrish2may
0

Step 1: Calculate moles of oxygen and nitrogen gas

Since we know \text PPP, \text VVV,and \text TTT for each of the gases before they're combined, we can find the number of moles of nitrogen gas and oxygen gas using the ideal gas law:

 

RT

PV

​  n, equals, start fraction, P, V, divided by, R, T, end fraction

Solving for nitrogen and oxygen, we get:

N  

2

​  

​  =  

(0.08206  

mol⋅K

atm⋅L

​  )(273K)

(2atm)(24.0L)

​  =2.14mol nitrogenn, start subscript, N, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, space, a, t, m, right parenthesis, left parenthesis, 24, point, 0, space, L, right parenthesis, divided by, left parenthesis, 0, point, 08206, space, start fraction, a, t, m, dot, L, divided by, m, o, l, dot, K, end fraction, right parenthesis, left parenthesis, 273, K, right parenthesis, end fraction, equals, 2, point, 14, space, m, o, l, space, n, i, t, r, o, g, e, n

O  

2

​  

​  =  

(0.08206  

mol⋅K

atm⋅L

​  )(273K)

(2atm)(12.0L)

​  =1.07mol oxygenn, start subscript, O, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, space, a, t, m, right parenthesis, left parenthesis, 12, point, 0, space, L, right parenthesis, divided by, left parenthesis, 0, point, 08206, space, start fraction, a, t, m, dot, L, divided by, m, o, l, dot, K, end fraction, right parenthesis, left parenthesis, 273, space, K, right parenthesis, end fraction, equals, 1, point, 07, space, m, o, l, space, o, x, y, g, e, n

Step 2 (method 1): Calculate partial pressures and use Dalton's law to get \text P_\text{Total}P  

Total

Once we know the number of moles for each gas in our mixture, we can now use the ideal gas law to find the partial pressure of each component in the 10.0\,\text L10.0L10, point, 0, space, L container:

\text P = \dfrac{\text{nRT}}{\text V}P=  

V

nRT

​  P, equals, start fraction, n, R, T, divided by, V, end fraction

N  

2

​  

​  =  

10L

(2.14mol)(0.08206  

mol⋅K

atm⋅L

​  )(273K)

​  =4.79atmP, start subscript, N, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 2, point, 14, space, m, o, l, right parenthesis, left parenthesis, 0, point, 08206, space, start fraction, a, t, m, dot, L, divided by, m, o, l, dot, K, end fraction, right parenthesis, left parenthesis, 273, space, K, right parenthesis, divided by, 10, space, L, end fraction, equals, 4, point, 79, space, a, t, m

2

​  

​  =  

10L

(1.07mol)(0.08206  

mol⋅K

atm⋅L

​  )(273K)

​  =2.40atmP, start subscript, O, start subscript, 2, end subscript, end subscript, equals, start fraction, left parenthesis, 1, point, 07, space, m, o, l, right parenthesis, left parenthesis, 0, point, 08206, space, start fraction, a, t, m, dot, L, divided by, m, o, l, dot, K, end fraction, right parenthesis, left parenthesis, 273, space, K, right parenthesis, divided by, 10, space, L, end fraction, equals, 2, point, 40, space, a, t, m

Notice that the partial pressure for each of the gases increased compared to the pressure of the gas in the original container. This makes sense since the volume of both gases decreased, and pressure is inversely proportional to volume.

We can now get the total pressure of the mixture by adding the partial pressures together using Dalton's Law:

P  

Total

​  

​    

=P  

N  

2

​  

​  +P  

O  

2

​  

​  

=4.79atm+2.40atm=7.19atm

​  

Step 2 (method 2): Use ideal gas law to calculate \text P_\text{Total}P  

Total

​  P, start subscript, T, o, t, a, l, end subscript without partial pressures

Since the pressure of an ideal gas mixture only depends on the number of gas molecules in the container (and not the identity of the gas molecules), we can use the total moles of gas to calculate the total pressure using the ideal gas law:

P  

Total

​  

​    

=  

V

(n  

N  

2

​  

​  +n  

O  

2

​  

​  )RT

​  

=  

10L

(2.14mol+1.07mol)(0.08206  

mol⋅K

atm⋅L

​  )(273K)

​  

=  

10L

(3.21mol)(0.08206  

mol⋅K

atm⋅L

​  )(273K)

​  

=7.19atm

​  

\text P_{\text {N}_2} = x_{\text N_2} \text {P}_{\text{Total}} =\left(\dfrac{2.14\,\text{mol}}{3.21\,\text{mol}}\right)(7.19\,\text{atm})=4.79\,\text {atm}P  

N  

2

​  

​  =x  

N  

2

​  

​  P  

Total

​  =(  

3.21mol

2.14mol

​  )(7.19atm)=4.79atmP, start subscript, N, start subscript, 2, end subscript, end subscript, equals, x, start subscript, N, start subscript, 2, end subscript, end subscript, P, start subscript, T, o, t, a, l, end subscript, equals, left parenthesis, start fraction, 2, point, 14, space, m, o, l, divided by, 3, point, 21, space, m, o, l, end fraction, right parenthesis, left parenthesis, 7, point, 19, space, a, t, m, right parenthesis, equals, 4, point, 79, space, a, t, m

\text P_{\text {O}_2} = x_{\text O_2} \text {P}_{\text{Total}} =\left(\dfrac{1.07\,\text{mol}}{3.21\,\text{mol}}\right)(7.19\,\text{atm})=2.40\,\text {atm}P  

O  

2

​  

​  =x  

O  

2

​  

​  P  

Total

​  =(  

3.21mol

1.07mol

Answered by alabhsh
0

Answer:

Explanation:we know

*PV:NRT*

As when two gases are in a container each gas behaves as if it is present alone in that container so each gas behaves independently

N1:3 moles

N2:2moles

P1:5atm

P2:?

In closed vessel

P is variable

V is constant

N Is constant

P1V1:NRT

5xv: 3x0.0821 XT ............1

P2v:nRt

P2v:2x0.0821xt..........2

Divide 2 by 1

P2/5:2/3

P2:10/3

Which is required

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