I know this is a larger question, I upped the points to the max I can! Please be
ID: 1307803 • Letter: I
Question
I know this is a larger question, I upped the points to the max I can! Please be as detailed as possible. Mainly, I am looking for what equations to apply to this. If I can get the equations, and what the variables within them represent, that would be sufficient. Thank you!!!
Questions:
20. A cylinder contains 25 mol of a monatomic ideal gas at a temperature of 25 degrees C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.013x105 Pa (1 atm) on the gas. The gas is heated until its temperature increases to 132 degrees C. What was the initial volume of the gas?
a. 0.711 m^3
b. 0.611 m^3
c. 0.0513 m^3
d. 0.0413 m^3
e. cannot answer without knowing the final volume
21. By how much did the volume increase or decrease?
a. increased by magnitude of [0.831 - answer in #20] m^3
b. decreased by magnitude of [0.831 - answer in #20] m^3
c. increased by magnitude of [0.271 - answer in #20] m^3
d. decreased by magnitude of [0.271 - answer in #20] m^3
e. cannot answer without knowing the final volume
22. How much work is done on or by the gas in this process?
a. (1.013x105 Pa) times (answer in #20)
b. (1 atm) times (answer in #20)
c. (1.013x105 Pa) times (answer in #21)
d. (1 atm) times (answer in #21)
e. n R T times ln(answer in #20 / answer in #21)
23. What is the change in thermal energy of the gas?
a. increased by 7794 J
b. increased by 33360 J
c. increased by 41154 J
d. increased by magnitude of answer in #22
e. no change in thermal energy
24. How much heat entered or left the gas?
a. 7794 J entered gas
b. 33360 J entered gas
c. 41154 J entered gas
d. (answer in #23) + (answer in #22)
e. (answer in #23) - (answer in #22)
Explanation / Answer
20) V2 = nRT2/P = 25*8.314*(132+273)/(1.0135*10^5) = 0.83 m3
V1/T1 = V2/T2
==> V1 = 0.83*(25+273)/(273+132) = 0.611 m3
correct option is b. 0.611 m^3
21) increase in volume = 0.83-0.61 =0.22
sp correct option is c. increased by magnitude of [0.271 - answer in #20] m^3
22) c. (1.013x105 Pa) times (answer in #21)
23)change in tehrmal enegy = n*Cv*(T2-T1)
Cv = 1.5*R = 12.5
change in energy = 25*12.5*(132-25) = 33437 J
so correct option is b. increased by 33360 J
24) heat entereded or left = change in themal energy - Work done
so correct option is e. (answer in #23) - (answer in #22)