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
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
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