The standard Gibbs free energy of formation ΔGf° of Rb(s), H2(g) and Pb(s) are all zero. Similar to enthalpies of formation, the values of the standard Gibbs energies of formation are zero for the elements in their most stable forms at room conditions 298 Kelvin and one atmosphere pressure.
what do you mean? explain what you r trying to dsay
Answer:
2100 kPa
Explanation:
The temperature is constant, so the only variables are pressure and volume.
We can use Boyle’s Law.
p₁V₁ = p₂V₂ Divide both sides of the equation by V₂
p₂ = p₁ × V₁/V₂
p₁ = 485 kPa; V₁ = 648 mL
p₂ = ?; V₂ = 0.15 L = 150 mL Calculate p₂
p₂ = 485 × 648/150
p₂ = 2100 kPa
T eg(1) : 97.5°C, T eg(2): 98.5°C, T eg(3): 99.2°C
∆T water(1): -2.5°C, ∆T water(2): -1.5°C, ∆T water(3): -0.8°C
∆T metal(1): 77.5°C, ∆T metal(2): 80.5°C, ∆T metal(3): 80.2°C.
ft= (m1 cp1 t1 + m2 cp2 t2 + .... + mn cpn tn) / (m1 cp1 + m2 cp2 + .... + mn cpn) (1)
where,
1000g = 1kg
ft(t eg)= final mixed temperature (°C)
m = mass of substance (kg)
cp = specific heat of substance (J/kg°C)
t = temperature of substance (°C)
