Answer:
The work done on neon = -323 J
The internal energy change= -392.84 J
The heat absorbed by neon = -69.84 J
Explanation:
Step 1: Data given
Number of moles = 0.500 moles
Pressure = 1 atm
Temperature = 273 Kelvin
The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.
a) calculate the work done on neon
W = -P(V2-V1)
⇒ with P = the pressure = 0.1 atm
⇒ with V1 = the initial volume = nRTi /Pi
⇒ with V2 = the final volume = nRTf /Pf
W = -PnR((T2/P2) -(T1/P1))
⇒ with T2 = the final temperature = 210 K
⇒ with T1 = the initial temperature = 273 K
⇒ with P2 = the final pressure = 0.200 atm
⇒ with P1 = the initial pressure = 1.00 atm
W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))
W = -nR*(105 - 27.3)
W= -(0.500)*(8.314)*(77.7)
W = -323 J
b) calculate the internal energy change
E = (3/2)*nRT
ΔE = Ef - Ei
ΔE =(3/2)*nR(T2-T1)
⇒ with n= number of moles = 0.500 moles
⇒ with T2 =the final temperature = 210 K
⇒ with T1 = the initial temperature = 273 K
ΔE = (3/2)*(0.5)*(8.314)(210-273)
ΔE = -392.84 J
c) Calculate the heat absorbed by neon
ΔE = q + W
q = ΔE -W
⇒ with ΔE = -392.84 J
⇒ with W = -323 J
q = -392.84 J -( -323 J)
q =-392.84 J + 323 J
q = -69.84 J
