When ΔG° is the change in Gibbs free energy
So according to ΔG° formula:
ΔG° = - R*T*(㏑K)
here when K = [NH3]^2/[N2][H2]^3 = Kc
and Kc = 9
and when T is the temperature in Kelvin = 350 + 273 = 623 K
and R is the universal gas constant = 8.314 1/mol.K
So by substitution in ΔG° formula:
∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)
= - 4536
Answer:
219.95 °C
Explanation:
Given data:
Volume of gas = 9.71 L
Initial pressure = 209 torr (209/760 = 0.275 atm)
Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)
Final temperature = ?
Final pressure = 364 torr (364/760 =0.479 atm)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
0.275 atm / 283.1 K = 0.479 atm/T₂
T₂ = 0.479 atm × 283.1 K/ 0.275 atm
T₂ = 135.6 atm. K /0.275 atm
T₂ = 493.1 K
Kelvin to °C:
493.1 K - 273.15 = 219.95 °C
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
energy is required to move from one state or phase of matter to the next. Energy is used to make a liquid into a gas or a solid into a liquid.
Https://www.youtube.com/watch?v=LXDG0nqYdR4