A 2.0-{\ m mol} sample of ideal gas with molar specific heat C_V = {\\textstyle{

Question

A 2.0-{ m mol} sample of ideal gas with molar specific heat C_V = { extstyle{5 over 2}},R is initially at 320 K and 100 { m kPa} pressure.

Part A

Determine the final temperature when 1.7 { m kJ} of heat are added to the gas isothermally.

Express your answer using two significant figures.

ANSWER:

T_f =

{ m K}

Part B

Determine the work done by the gas when 1.7 { m kJ} of heat are added to the gas isothermally.

Express your answer using two significant figures.

ANSWER:

W =

{ m kJ}

Part C

Determine the final temperature when 1.7 { m kJ} of heat are added to the gas at constant volume.

Express your answer using two significant figures.

ANSWER:

T_f =

{ m K}

Part D

Determine the work done by the gas when 1.7 { m kJ} of heat are added to the gas at constant volume.

Express your answer using two significant figures.

ANSWER:

W =

{ m J}

Part E

Determine the final temperature when 1.7 { m kJ} of heat are added to the gas isobarically.

Express your answer using two significant figures.

ANSWER:

T_f =

{ m K}

Part F

Determine the work done by the gas when 1.7 { m kJ} of heat are added to the gas isobarically.

Express your answer using two significant figures.

ANSWER:

W =

{ m J}

Explanation / Answer

part A

as iso thermally

delta T =0

Tf= Ti = 320K

Part B

work done by gas + change in internal energy = heat added

as delta U =0

hence work done

= 1.7 kJ

Part C

at constant volume Cv= 5/ 2R

delta U = nCv*delta T = 1.7kJ

n=2

we get delta T= 40.895

Tf = 360.895 K

Part D

at constant V work done is 0

Part E

Cp = Cv +R

= 7R/2

nCp delta T = 1.7kJ

delta T = 29.21

we get

Tf= 349.21 K

Part F

work done = nCp*delta T - nCv *delta T

we get

= n*R*delta T

=485.70 J

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