20.9 J heat ideal gas 50.0 cm3 to 100.0 cm3 at 1.00 atm

20.9 J heat ideal gas 50.0 cm3 to 100.0 cm3 at 1.00 atm

 

 

When 20.9 J was added as heat to a particular ideal gas, the volume of the gas changed from 50.0 cm3 to 100.0 cm3 while the pressure remained at 1.00 atm.

(a) By how much did the internal energy of the gas change? If the quantity of gas present was 2.00×10^(-3) mol, find (b) Cp and (c) Cv.

 

------------------------

 

V1 = (50.0 cm3) (1 mL / 1 cm3) (1 L / 1000 mL)

= (50.0) (1 / 1000)

= 0.0500 L

 

V2 = (100.0) (1 / 1000)

= 0.1000 L

 

 

 

일정 압력 조건이므로,

w = –PΔV

( 참고 https://ywpop.tistory.com/5281 )

 

= –(1.00 atm) (0.1000 – 0.0500 L)

= –(1.00) (0.1000 – 0.0500)

= –0.0500 atm•L

 

 

(–0.0500 atm•L) (8.314 J / 0.08206 atm•L)

= (–0.0500) (8.314 / 0.08206)

= –5.067 J

 

 

 

열역학 제1법칙

ΔE = q + w 또는

ΔU = q + w

( 참고 https://ywpop.tistory.com/5134 )

 

= (+20.9 J) + (–5.067 J)

= 15.8 J

 

 

300x250

 

 

PV = nRT 로부터,

 

T1 = PV1 / nR

= [(1.00 atm) (0.0500 L)] / [(2.00×10^(-3) mol) (0.08206 atm•L/mol•K)]

= [(1.00) (0.0500)] / [(2.00×10^(-3)) (0.08206)]

= 304.66 K

 

T2 = PV2 / nR

= [(1.00) (0.1000)] / [(2.00×10^(-3)) (0.08206)]

= 609.31 K

 

 

 

q_p = nCpΔT

( 참고 https://ywpop.tistory.com/23639 )

 

Cp = q_p / nΔT

= 20.9 J / [(2.00×10^(-3) mol) (609.31 – 304.66 K)]

= 20.9 / [(2.00×10^(-3)) (609.31 – 304.66)]

= 34.3 J/mol•K

 

 

 

ΔU = nCvΔT

 

Cv = ΔU / nΔT

= 15.8 / [(2.00×10^(-3)) (609.31 – 304.66)]

= 25.9 J/mol•K

 

 

 

답:

(a) ΔU = 15.8 J

(b) Cp = 34.3 J/mol•K

(c) Cv = 25.9 J/mol•K

 

 

 

 

[참고] Cp – Cv = R

= 34.3 – 25.9

= 8.4 ≒ 8.314

 

 

 

 

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