Exercise 19.22 Three moles of an ideal monatomic gas expand at a constant pressu

Question

Exercise 19.22 Three moles of an ideal monatomic gas expand at a constant pressure of 3.00 atm ; the volume of the gas changes from 3.20×102 m3 to 4.20×102 m3 . Part A Calculate the initial temperature of the gas. Tinitial = K Part B Calculate the final temperature of the gas. Tfinal = K Part C Calculate the amount of work the gas does in expanding. W = J Part D Calculate the amount of heat added to the gas. Q = J Part E Calculate the change in internal energy of the gas. U = J Exercise 19.22 Three moles of an ideal monatomic gas expand at a constant pressure of 3.00 atm ; the volume of the gas changes from 3.20×102 m3 to 4.20×102 m3 . Part A Calculate the initial temperature of the gas. Tinitial = K Part B Calculate the final temperature of the gas. Tfinal = K Part C Calculate the amount of work the gas does in expanding. W = J Part D Calculate the amount of heat added to the gas. Q = J Part E Calculate the change in internal energy of the gas. U = J Exercise 19.22 Three moles of an ideal monatomic gas expand at a constant pressure of 3.00 atm ; the volume of the gas changes from 3.20×102 m3 to 4.20×102 m3 . Part A Calculate the initial temperature of the gas. Tinitial = K Part B Calculate the final temperature of the gas. Tfinal = K Part C Calculate the amount of work the gas does in expanding. W = J Part D Calculate the amount of heat added to the gas. Q = J Part E Calculate the change in internal energy of the gas. U = J

Explanation / Answer

part A
Use ideal gas law
PV = nRT
=>
T = PV / (nR)
= 3 101325 Pa 3.2×10² m³ / ( 3 mol 8.314472 Pam²K¹mol¹)
= 389.98 K


part B
From ideal gas follows that
V/T = nR/P = constant
for a constant pressure process.
Hence,
V/T = V/T
=>
T = T ( V/V)
= 389.98 K ( 4.2×10² m³ / 3.2×10² m³)
= 511.84 K


part C
Work done by the gas is given by the integral:
W = P dV from initial to final volume
For a constant pressure process this simplifies to:
W = P dV = PV
Hence,
W = P(V - V)
= 3 101325 Pa (4.2×10² m³ - 3.2×10² m³)
= 3039.76 Pam³
= 3039.76vJ


part D

Heat transferred in a constant pressure process equals the change in enthalpy:
Q = H
Change in enthalpy for an ideal gas is given by:
H = nCpT
The molar heat capacity at constant pressure for a monatomic ideal gas is:
Cp = (5/2)R
Hence,
Q = (5/2)nR(T - T)
= (5/2) 3 mol 8.314472 JK¹mol¹ (511.84 K - 389.98 K)
= 7598.59 J


part E
Change in internal energy equals heat added to minus work done by the gas:
U = Q - W
= 7598.59 J - 3039.76 J
= 4558.84 J

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