States of Matter
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5 months agoContributor-Level 10
5.12. Given,
P= 3.32 bar
V= 5 dm3
n= 4 mol
R= 0.083 bar dm3 K-1 mol-1
PV = nRT
Or T = PV / nR = 3.32 x 5 dm3 / (4.0 mol x 0.083 bar dm3 K-1 mol-1)
= 50 K
New answer posted
5 months agoContributor-Level 10
5.10. We know,
PV=nRT
n = PV/RT
n= (0.1 x34.05 x 10-3)/ (0.083 x 819)
&n
New question posted
5 months agoNew answer posted
5 months agoContributor-Level 10
5.8. Calculation of partial pressure of H2 in 1L vessel:
P1= 0.8 bar, P2=? V1= 0.5 L, V2 = 1.0 L
As temperature remains constant, P1V1 = P2V2
=> (0.8 bar) (0.5 L) = P2 (1.0 L)
=>P2 = 0.40 bar, i.e., PH2 = 0.40 bar
Calculation of partial pressure of O2 in 1 L vessel
P1V1 = P2V2
(0.7 bar) (2.0 L) = P2 (1L)
=>P2 = 1.4 bar
=>Po2= 1.4 bar
Total pressure =PH2 + PQ2 = 0.4 bar + 1.4 bar = 1.8 bar
New answer posted
5 months agoContributor-Level 10
5.7. Applying PV = nRT
We get P = nRT / V
Given: nCH4 = 3.2 / 16 mol = 0.2 mol
nCO2 = 4.4 /44 mol = 0.1 mol
So,
PCH4 = (0.2 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.55 atm
PCO2= (0.1 mol) (0.0821 dm3atm K-1 mol-1) (300 K) / (9 dm3)
= 0.27 atm
Ptotal =
New answer posted
5 months agoContributor-Level 10
5.6. The chemical equation for the reaction is
2 Al + 2 NaOH + H2O? 2 NaAlO2 + 3H2
i.e. 2 moles of Al give 3 moles of H2 gas.
Moles of aluminium = 270.15g = 5.56*10?3 moles
Moles of H2 produced= 23*5.56*10?3
= 8.33*10?3 moles
Volume of H2 = nRT / P
&nbs
New answer posted
5 months agoContributor-Level 10
5.5. Suppose molecular masses of A and B are MA and MB respectively. Then their number of moles will be
nA= 1/MA nB = 2/MB
Given: PA = 2 bar and PA + PB = 3 bar
=> PB = 1bar
Since, PV = nRT
PA= nART and PBV= nBRT
Therefore, (PA / PB) = (nA / nB)= (MB / 2MA)
=> MB / MA = 2 x PA / PB = 2 x 2 /1 = 4
=> MB = 4 MA
New answer posted
5 months agoContributor-Level 10
Answer: According to ideal gas equation
PV = nRT
Or P= nRT/V
Replacing n by m/M, we get
P M1 x 2 = 28 x 5 (Molecular mass of N2 = 28 g/mol)
or
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5 months agoNew question posted
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