2. consider a process where propane gas, which is initially at 1 bar and 35 ºc, is compressed to a final state of 135 bar and 195 ºc. estimate the molar volume of the gas in its final state. estimate also the entropy and enthalpy changes for this process. you can assume that propane behaves as an ideal gas in the initial state.

This is an Ideal Gas Process. The Ideal gas equation states that:

PxV=nxRxT

where n = number of moles of the gas

P = pressure of the gas at any state

V= Volume of the gas at any state

T=Temperature of the gas at any state

Initial State :  P₁ = 1Bar=1.00x10⁵N/m², T₁=35°C = 308K,    V₁=?

Final State:    P₂=135Bar=1.35x10⁷N/m², T₂=195°C=468K, V₂=?

From Eqn 1, we can calculate the molar volume of the Gas at the final state Q=n/V₂ =P₂/RT₂                                     . .  .  . . . . . .             Eqn 1

Q=1.35x10⁷N/m²/(8.314J/mol-K x 468K)

Q=33,828mol/m³ = 33.8Kmol/m³

The molar volume of the Gas at the final state Q =33.8Kmol/m³

To estimate the Enthalpy and the Entropy change for an ideal gas process requires these equations which are derived from the law of conservation of energy (1st and 2nd law of thermodynamics):

ΔH = CpΔT

ΔS = CpInΔT - RInΔP/M

ΔT=T2-T1 and ΔP=P2-P1

Cp=Specific heat capacity of an ideal propane gas = 1.68kJ/KgK

R = Universal gas constant = 8.314J/molK

For Enthalpy change, ΔH = 1.68 kJKgK x (468 - 308)K

ΔH = 268.8KJ/kg

For Entropy change, ΔS = (CpInΔT) - (RInΔP)/M      M=Propane molar mass=44.1kg/mol

ΔS=(1.68xln(468/308)) - (8.314xln(135/1)/44.1)

ΔS   =0.703KJ/KgK - 0.925KJ/kgK

 ΔS = - 0.222KJ/KgK

Therefore, the entropy and enthalpy changes for this process are - 0.222KJ/KgK and 268.8KJ/kg.

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

c)red bank

explanation:

red bank is a borough in monmouth county, new jersey, incorporated in 1908 and located on the navesink river, the area's original transportation route to the ocean and other ports.