Question

A cylinder of volume 0.240 m³ contains 10.1 mol of neon gas at 21.6°C. Assume neon behaves as an ideal gas

(g) Compute ΔU − W.

(i) Compute Q, the thermal energy transfer.

Answer 1

(a)
Use ideal gas law:
P∙V = n∙R∙T
=>
P = n∙R∙T/V
= 10.1 mol ∙ 8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ (21.6 + 273) K / 0.240 m³
=103081Pa
= 1.03081 atm

(b)
Internal energy for an ideal gas is given by:
U = n∙Cv∙T
The molar heat capacity for a monatomic ideal gas like neon is:
Cv = (3/2)∙R
Hence,
U = (3/2)∙n∙R∙T
= (3/2) ∙ 10.1 mol ∙ 8.3145 J∙K⁻¹∙mol⁻¹ ∙ (21.6+ 273) K
= 37109.2 J
= 37.109 kJ

(c)
Assuming reversible operation the work done on the gas is given by the integral
W = - ∫ P dV from initial to final volume
For a constant pressure process this simply fies to:
W = - P ∙ ∫ dV = - P∙∆V
=>
W = - 103081 Pa ∙ (1.00 m³ - 0.24 m³ )
= - 78341.6 J
= - 78.3416 kJ

(d)
When you separate variable and constant terms in ideal gas law equation you find:
V/T = n∙R/P = constant
=>
V_final/T_final = V_initial/T_initial
=>
T_final = T_initial ∙ (V_final/V_initial)
= (21.6 + 273)K ∙ (1.00 m³ / 0.24 m³)
= 1227.5K
= 954.5 °C

(e)
U = (3/2)∙n∙R∙T
= (3/2) ∙ 10.1 mol ∙ 8.3145 J∙K⁻¹∙mol⁻¹ ∙ 1227.5 K
= 154.62 kJ

(f)
∆U = U_final - U_initial
= 154.62 kJ - 37.109 kJ
=117.512 kJ

(g)
∆U-W = 117.512-(-78.341) = 195.853 KJ

(h)
The internal energy of the gas increases although it does work, i.e. it transfer energy to its surroundings. So there must be an additional energy transfer to the gas, such you get an net positive change in internal energy.

(i)
The heat transferred in a constant pressure process equals change in enthalpy:
Q = ∆H
with the formula for enthalpy of an ideal gas
H = n∙Cp∙T
and the molar heat capacity for a monatomic ideal gas
Cp = (5/2)∙R
you get
Q = (5/2)∙n∙R∙∆T
= (5/2) ∙ 10.1mol ∙ 8.3145 J∙K⁻¹∙mol⁻¹ ∙ (1227.5 K + 294.6 K )
= 319.55 kJ