Question

Problem 19.54

A cube 21 cm on each side contains 3.2 g of helium at 20∘C. 1200 J of heat energy are transferred to this gas.

Part A

What is the final pressure if the process is at constant volume?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the final volume if the process is at constant pressure?

Express your answer to two significant figures and include the appropriate units.

Answer 1

n = m/M = 3.2g / 4g/mol = .8mol
of helium in a cube of initial volume
V₀ = a³ = (18cm)³ = 9261cm³ = 9.261L = 9.2612×10⁻³m³

Initial pressure can be found using ideal gas law:
P₀ = n∙R∙T₀/V₀
= .8mol ∙ 8.314472J/molK ∙ (20 + 273.15)K / 9.261×10⁻³m³
= 210550.694Pa
= 2.1055atm

(i) constant volume process
The change of pressure can be found from ideal gas
P = n∙R∙T/V
because n and V are constant:
∆P = (n∙R/V)∙∆T

The heat transferred in a constant volume process is equal to the change of internal energy
Q = ∆U = n∙Cv∙∆T
Molar heat capacity at constant volume of a monatomic ideal gas, like helium, is
Cv = (3/2)∙R
So temperature change and heat transferred are related as:
∆T = (2/3)∙Q / (n∙R)

Hence:
∆P = (n∙R/V) ∙ (2/3)∙Q / (n∙R) = (2/3)∙Q / V
= (2/3) ∙ 1200J / 9.2612×10⁻³m³
= 83229.3Pa
=>
P_final = P₀+ ∆P = 210550.694Pa + 83229.3Pa
= 293779.994Pa
= 2.94atm


(ii) constant pressure process
The change of volume can be found from ideal gas
V = n∙R∙T/P
because n and p are constant:
∆V = (n∙R/P)∙∆T

The heat transferred in a constant pressure process is equal to the change of enthalpy
Q = ∆H = n∙Cp∙∆T
Molar heat capacity at constant pressure of a monatomic ideal gas, like helium is
Cp = (5/2)∙R
So temperature change and heat transferred are related as:
∆T = (2/5)∙Q / (n∙R)

Hence:
∆V = (n∙R/P) ∙ (2/5)∙Q / (n∙R) = (2/5)∙Q / P
= (2/5) ∙ 1200J / 210550.694Pa
= 2.28×10⁻³m³
= 2.28L
=>
V_final = V₀+ ∆V = 9.2612×10⁻³m³ + 2.28×10⁻³m³
= 11.9×10⁻³m³
= 11.9L