Answer:
Part (a) 0.100 mole of He(g) at 258°C and a volume of 5.00 L, entropy = 12.6117 J/K
Part (b) 3.00 moles of He(g) at 258°C and a volume of 3000.0 L, entropy = 379.089 J/K
Explanation:
ΔS = n*R*ln(V₂/V₁)
where;
ΔS is change in entropy
n is number of moles
R is ideal gas constant = 0.08206 L*atm/mol*K
V₂ is final volume
V₁ is the initial volume
Part (a) 0.100 mole of He(g) at 258°C and a volume of 5.00 L
Initial volume of 0.1 mole of helium gas at 258°C (495K)
V₁ = n*R*T/P = (0.100mol)(0.08206 L*atm/mol*K)*(495K)/(1atm) = 4.06 L
ΔS = n*R*ln(V₂/V₁) = 0.1*0.08206*ln(5/4.06) = 0.0017 J/K
ΔS = 0.0017 J/K
S₁ = 0.100 mol* 126.1(J/K*mol) = 12.61 J/K
S₂ = ΔS + S₁
S₂ = 0.0017 J/K + 12.61 J/K = 12.6117 J/K
Part (b) 3.00 moles of He(g) at 258°C and a volume of 3000.0 L
Initial volume of 3.00 moles of helium gas at 258°C (495K)
V₁ = n*R*T/P = (3.00 mol)(0.08206 L*atm/mol*K)*(495K)/(1atm) = 121.8591 L
ΔS = n*R*ln(V₂/V₁) = 3.0*0.08206 *ln(3000/121.859) = 0.789 J/K
ΔS = 0.789 J/K
S₁ = 3.0 mol* 126.1(J/K*mol) = 378.3 J/K
S₂ = ΔS + S₁
S₂ = 0.789 J/K + 378.3 J/K = 379.089 J/K
The entropy is the following:
(a) entropy is 12.6117 J/K
(b) entropy is 379.089 J/K
Entropy is the measurement of the thermal energy of a system that is unavailable for use.
[tex]\Delta S = n\times R \times ln \dfrac{V_2}{V_1}[/tex]
The change in entropy is ΔS
Number of moles is n
The ideal gas constant is R = 0.08206 L x atm/mol x K
Final volume is V₂
The initial volume is V₁
(a) 0.100 mole of He(g) at 258°C and a volume of 5.00 L
Initial of 0.1 mole of helium gas at 258°C (495K)
[tex]\rm V_1 = n\times R\times \dfrac{T}{P} = (0.100mol)(0.08206 \times atm/mol\times K)\times dfrac{(495K)}{(1atm)}= 4.06 L[/tex]
[tex]\Delta S = 0.1\times0.08206 \times ln \dfrac{5}{4.06} = 0.0017 J/K[/tex]
[tex]S_1 = 0.100 mol \times 126.1(J/K\times mol) = 12.61\; J/K[/tex]
S₂ = ΔS + S₁
S₂ = 0.0017 J/K + 12.61 J/K = 12.6117 J/K
(b) 3.00 moles of He(g) at 258°C and a volume of 3000.0 L
Initial volume of 3.00 moles of helium gas at 258°C (495K)
[tex]\rm V_1 = n\times R\times \dfrac{T}{P} = (3.00mol)(0.08206 \times atm/mol\times K)\times dfrac{(495K)}{(1atm)}= 121.8591\; L[/tex]
[tex]\rm \Delta S = 3.0\times0.08206 \times ln \dfrac{3000}{121.859} = 0.789 J/K[/tex]
[tex]\rm S_1 = 3.0 mol \times 126.1(J/K\times mol) = 378.3\; J/K[/tex]
S₂ = ΔS + S₁
S₂ = 0.789 J/K + 378.3 J/K = 379.089 J/K
Thus, The entropy is a- 12.6117 J/K
b- 379.089 J/K
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