When plotting a single AC cycle beginning at zero degrees and moving forward in time the cycles negative peak occurs at

Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states

Vadim26 [7]

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

4 0

3 years ago

Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate

uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae $^{-E/RT}$ ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e $^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}$

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

3 0

3 years ago

The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

madam [21]

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss :Due to friction of pipe surface

2.Minor loss :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

$Losses=K\dfrac{V^2}{2g}$