Answer:
V1=5<u>ft3</u>
<u>V2=2ft3</u>
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5<u>ft3</u>
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= <u> 2ft3</u>
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
Answer:

Explanation:
First we calculate the mass of the aire inside the rigid tank in the initial and end moments.
(i could be 1 for initial and 2 for the end)
State1


State2


So, the total mass of the aire entered is

At this point we need to obtain the properties through the tables, so
For Specific Internal energy,

For Specific enthalpy

For the second state the Specific internal Energy (6bar, 350K)

At the end we make a Energy balance, so

No work done there is here, so clearing the equation for Q



The sign indicates that the tank transferred heat<em> to</em> the surroundings.
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- 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)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
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))