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Secure https: session, masteringphysics.com 73043464 PHYS 212 Sp17 C19 Problem 1

ID: 1524265 • Letter: S

Question

Secure https: session, masteringphysics.com 73043464 PHYS 212 Sp17 C19 Problem 19.3 Problem 19.31 Part A 0.10 mol of nitrogen gas follow the two processes shown in the figure(Figure 1) How much heat is required for process A? Express your answer to two significant figures and include the appropriate units. Value Units QA Submit My Answers Give Up incorrect; Try Again Part B How much heat is required for process B? Express your answer to two significant figures and include the appropriate units. Figure 1 of 1 QB Value Units p (atm) Submit My Answers Give Up incorrect Try Again v (cm") 1000 2000 3000 He previous l 10 of 12 l next Provide Feedback Continue

Explanation / Answer

A
Process A is a constant volume process. The heat transferred in such a process equals the change in internal energy of the gas.
Q = U

Assuming ideal gas behavior the change in internal energy of the nitrogen gas is given by:
U = nCvT
The molar heat capacity at constant volume for an diatomic ideal gas like nitrogen is
Cv = (5/2)R
Hence,
Q =(5/2)nRT

From ideal gas law
nRT = pV
follows for an constant volume process:
nRT = (pV) = Vp

Hence,
Q =(5/2)Vp = (5/2)V(p_f - p_i)
= (5/2) 2000cm³ (1atm - 3atm)
= (5/2) 2000m³×10m³ (- 2 1.01325×10 Pa)
= -1013.25 J


B
Process B is a constant pressure process. The heat transferred in such a process equals the change in enthalpy of the gas.
Q = H

Assuming ideal gas behavior the change in enthalpy of the nitrogen gas is given by:
H = nCpT
The molar heat capacity at constant pressure for an diatomic ideal gas like nitrogen is
Cv = (7/2)R
Hence,
Q =(5/2)nRT

From ideal gas law
nRT = pV
follows for an constant pressure process:
nRT = (pV) = pV

Hence,
Q =(7/2)pV = (5/2)p(V_f - V_i)
= (7/2) 2atm (3000cm³ - 1000cm³)
= (7/2) 21.01325×10 Pa 2000m³×10m³
= 1418.55 J

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