PHYSICS CHAPTER 19-21 WORKSHEET ***PLEASE PROVIDE DETAILS SOLUTIONS AND EXPLANAT

Question

PHYSICS CHAPTER 19-21 WORKSHEET

***PLEASE PROVIDE DETAILS SOLUTIONS AND EXPLANATIONS TO THE FOLLOWING!

***PLEASE ANSWER AS MANY QUESTIONS AS POSSIBLE. I ONLY HAVE A FEW QUESTIONS LEFT FOR THIS MONTH!!!

7. An ideal gas is at a pressure 1.00 * 10^(5) N/m^(2) and occupies a volume 2.00m^(3). If the gas is compressed to a volume 1.00 m^(3) while the temperature remains constant, what will be the new pressure in the gas?

8. How many moles of water (H20) molecules are in a 4.00m^(3) container at a pressure 8.00 * 10^(5) N/m^(2) and temperature 600degreecelsius? The ideal gas constant is R = 8.314 J/mol . K = 0.0821 L . atm/mol.K.

9. During an adiabatic process, 20 moles of a monatomic ideal gas undergo a temperature change from 450 K to 320 K starting from an initial pressure is 400 kPa. The ideal gas constant is R = 8.314 J/mol. K

a) What is the final volume of the gas?

b) How much heat does the gas exchange during this process?

c) What is the change in the internal (thermal) energy of the gas during this process?

***PLEASE NEED ANSWERS ASAP!!!

Explanation / Answer

Q7
Given Initial Pressure(P1)=1.00 * 10^5 N/m^2
Final Pressure(p2)=?
Initial Volume(v1)=2.00 m^3
Final Volume(V2)=1.00 m^3
Using Boyle's Law
P1V1= P2V2
P2= (P1*V1)/V2
= (1.00 * 10^5 N/m^2 * 2.00 m^3)/1.00 m^3
= 2 * 10^5 N/m^2

Q8
Given Volume (V)= 4.00m^(3)
Pressure (P)= 8.00 * 10^(5) N/m^(2)
Temperature (T)= 600 degree celsius = 600+273K= 873 K
Ideal gas constant(R) = 8.314 J/mol .
Number of Moles (n)=?
Using Ideal Gas Equation
PV =nRT
n= PV/RT
= (8.00 * 10^(5) N/m^(2) * 4.00m^(3))/(8.314 J/mol * 873 K)
= 0.004408 * 10^5 mol
= 440 moles

Q9
a)Given moles(n)=20 mol
Initial Temperature(T1) = 450 K
Final Temperature(T2) = 320 K
Initial pressure(P1) =400 kPa
Ideal gas constant is R = 8.314 J/mol
Y for monoatomic gas = 1.67
Initial Volume V1=?
Final Volume V2=?
Using Ideal Gas Equation
PV=nRT
For revisable adiabatic change Pressure and Volume are relelated as:
PV^Y = C where C is constant
Replacing Pressure from Ideal Gas Equation we have

(nRT/V) * V^Y = C
V^(Y-1) T = C/nR (constant)

Hence
(V1/V2)^(Y-1)= T2/T1
V2= ((T1/T2)^(1/(Y-1)))*v1 ----- =n (1)

Also V1= nRT1/P1
= (20 mol * 8.314 J/mol * 450 K)/ 400 kPa
= 187.065 m^3
Now From =n (1),
v2 = 187.065 m^3 * (450K / 320K)^(1/(1.67 - 1)) = 311 m^3

b) I think Heat Change is Zero in Adiabatic Process (Please confirm it)

c) Internal Energy change is Work Done in the process
and is given as Pressure * Change In Volume
= 400 kPa * (311-187) ^3
= 400 kPa * 124 m^3
= 49600 * 10^3 * 10^(-6)
= 49.6 J

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