Question

Review Part A The figure(Figure 1) shows a thermodynamic process followed by 110 ng of helium. You may want to review (Page 515) For help with math skills, you may want to review Conversion Factors Conversion Between Celsius and Kelvin Rearrangement of Equations Involving Multiplication and Determine the pressure (in atm) of the gas at points 1, 2, and 3 View Available Hint(s) Pl.P2iPs-0.999,5.0,0.999 atm Figure 1 of 1 SubmitP evious Answe Incorrect; Try Again; 9 attempts remaining Isotherm Part B Determine the temperature (in C) of the gas at points 1, 2, and 3 Pi View Available Hint(s) 133 C 1000 cm

Answer 1

m = given mass = 110 mg = 0.110 g

M = molar mass = 4 g/mol

n = number of moles

n = m/M = 0.110/4 = 0.0275

Part A)

At Point 1 :

T₁ = Temperature at point 1 = 133 C = 133 + 273 = 406 K

V₁ = Volume at point 1 = 1000 cm³ = 10^-3 m³

P₁ = pressure at point 1 = ?

Using the equation

P₁ V₁ = n R T₁

inserting the values

P₁ (10^-3 ) = (0.0275) (8.314) (406)

P₁= 92825.8 Pa = 0.916 atm

At point 2 :

P₂= 5 P₁ = 5 (0.916) = 4.58 atm

At point 3 :

P₃ = P₁ = 0.916 atm

Part B)

T₁ = 133 ^oC

For process 1 to 2 :

P₁/T₁ = P₂/T₂                               (Since Volume is constant)

P₁/(133 + 273) = 5P₁/T₂

T₂ = 2030 K

T₂= 2030 - 273 = 1757 ^oC

T₃ = T₂ = 1757 ^oC                                      (Since process is isothermal from 2 to 3 )

Part C)

V₁ = 1000 cm³

V₂= V₁ = 1000 cm³

V₁/T₁ = V₃ /T₃

1000/(133 + 273) = V₃ /2030

V₃ = 5000 cm³

Part D)

for process 1-2 :

$\Delta$V = 0

W_1-2= P₁$\Delta$V = 0

for process 2-3 :

W_2-3 = n R T₂ log(V₃/V₂)

W_2-3= (0.0275) (8.314) (2030) log(5000/1000) = 747 J

for process 3-1:

W_3-1= P₁ (V₁ - V₃) = (92825.8) (0.001 - 0.005) = - 371.3 J