Question

A 10 cm -diameter cylinder contains argon gas at 10 atm pressure and a temperature of 60 ∘C . A piston can slide in and out of the cylinder. The cylinder's initial length is 19 cm . 2600 J of heat are transferred to the gas, causing the gas to expand at constant pressure. 1-What is the final temperature of the cylinder? 2-What is the final length of the cylinder?

Answer 1

(a)
Heat transferred is an constant pressure process is equal to the change of enthalpy.
Change of enthalpy of an ideal gas is given by
?H = n?cp_m??T = n?cp_m?(T_final - T_initial)

for any ideal gas
cp,m - cv;m = R
so for a monatomic ideal gas
cp,m = (5/2)?R

Q = n?(5/2)?R?(T_final - T_initial)
=>
T_final = T_initial + Q/(n?(5/2)?R)

The number of moles can be found from ideal gas law
V_initial = (pi/4)?D²?L
= (pi/4)? (0.13m)²?0.19m = 0.00252 m³

n = p ?V_initial / (R ? T_initial )
= 10?101325Pa ? 0.00252m³ / (8.3145J/molK ? (60 + 273)K)
n = 0.922 mol

=>
T_final = 333K + 2600J/(0.922mol ? (5/2) ? 8.3145J/molK)
T_final = 468.66 K = 195.66^oC

b)
Because number of moles in the cylinder and pressure dont change:
V/T = n?R/p = constant
=>
V_initial / T_initial = V_final/T_final
<=>
(pi/4)?D²?L_initial / T_initial = (?/4)?D²??L_final / T_final
=>
L_final = L_initial ? (T_final / T_initial)
= 19cm ? (468.66K / 333K)
L = 26.74 cm