For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
Answer:
Partial pressure of nitrogen gas is 0.98 bar.
Explanation:
According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.
where,
= total pressure = 3.9 bar
= partial pressure of nitrogen gas
= partial pressure of oxygen gas
= partial pressure of argon gases
= Mole fraction of nitrogen gas = 0.25
= Mole fraction of oxygen gas = 0.65
= Mole fraction of argon gases = 0.10
Partial pressure of nitrogen gas :
Partial pressure of oxygen gas :
Partial pressure of argon gas :
