compare the pressure exerted by one mole of co2 on the wall of 2.5 litre vessel at 27°c applying
Answers
Answer:
(a) 5 moles of CO
2
in 1l vessel of 47°C.
n=5,v=1l,t=273.15+47=320.15k
vanderwaal's equation of state, a=3.592
b=0.0427
(p+a
v
2
n
2
)(V−nb)=nrt
(p+3.592×
1
2
5
2
)(1−5×0.0427)=5×0.0821×320.15
(p+89.8)=167.1
Pressure,P=77.297atm
(b) Volume occupied by CO
2
gas is negligible (b=0).
vander waal's equation for 1 mole of gas is, (p+
v
2
a
)(v−b)=nrt⟶(i)
(p+
v
2
a
)(v)=nrt
P=
v
RT
−
v
2
a
Using R=0.0821,T=273K,v=22.4litre for 1 mole of an ideal gas.
P=
22.4
0.0821×273
−
22.4
2
3.592
=0.9934atm
Explanation:
hope it helps
(a) 5 moles of CO
2
in 1l vessel of 47°C.
n=5,v=1l,t=273.15+47=320.15k
vanderwaal's equation of state, a=3.592
b=0.0427
(p+a
v
2
n
2
)(V−nb)=nrt
(p+3.592×
1
2
5
2
)(1−5×0.0427)=5×0.0821×320.15
(p+89.8)=167.1
Pressure,P=77.297atm
(b) Volume occupied by CO
2
gas is negligible (b=0).
vander waal's equation for 1 mole of gas is, (p+
v
2
a
)(v−b)=nrt⟶(i)
(p+
v
2
a
)(v)=nrt
P=
v
RT
−
v
2
a
Using R=0.0821,T=273K,v=22.4litre for 1 mole of an ideal gas.
P=
22.4
0.0821×273
−
22.4
2
3.592
=0.9934atm