Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
The pressure (in atm) in a 5 L tank with 37.50 grams of oxygen gas at 350 K is 6.715 atm.
<h3>How do we calculate the pressure of gas?</h3>
Pressure of gas will be calculated by using the ideal gas equation as:
PV = nRT, where
P = pressure = ?
V = volume = 5 L
R = universal gas constant = 0.082 L.atm / K.mol
T = temperature = 350 K
n is the moles of gas and it will be calculated as:
n = W/M, where
W = given mass = 37.50 g
M = molar mass of oxygen = 32g/mol
n = 37.50/32 = 1.17 mol
On putting all these values, we get
P = (1.17)(0.082)(350) / (5) = 6.715 atm
Hence resultant pressure of the oxygen gas is 6.715 atm.
To know more about ideal gas equation, visit the below link:
brainly.com/question/1056445
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Answer:
High surface tension causes boiling point to increase.
Explanation:
The surface tension is related to the psychical forces between the molecules. Increasing the surface tension will increase the amount of energy (heat) needed to take the molecules apart and go from liquid state to vapor state, so the boiling point of the particular liquid will increase.
Answer:
A and C are true , B and D are false
Explanation:
For A)
from the first law of thermodynamics (in differential form)
dU= δQ - δW = δQ - PdV
from the second law
dS ≥ δQ/T
then
dU ≤ T*dS - p*dV
dU - T*dS + p*dV ≤ 0
from the definition of Gibbs free energy
G=H - TS = U+ PV - TS → dG= dU + p*dV + V*dp - T*dS - S*dT
dG - V*dp + S*dT = dU - T*dS + p*dV ≤ 0
dG ≤ V*dp - S*dT
in equilibrium, pressure and temperature remains constant ( dp=0 and dT=0). Thus
dG ≤ 0
ΔG ≤ 0
therefore the gibbs free energy should decrease in an spontaneous process → A reaction with a negative Gibbs standard free energy is thermodynamically spontaneous under standard conditions
For B) Since the standard reduction potential is related with the Gibbs standard free energy through:
ΔG⁰=-n*F*E⁰
then, when ΔG⁰ is negative , E⁰ is positive and therefore a coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous.