explain An ideal gas initially at 330 K urdergoes an isobaric expansion at 2.50

Question

explain

An ideal gas initially at 330 K urdergoes an isobaric expansion at 2.50 kpa. The volume increases from 1.00 min to 3.00 m and 12.2 k is transferred to the gas by heat, (a) what is the change in internal energy of the gas? (b) what is the final temperature af the gas? Need Help? Rendia M One mole of an ideal gas does 3 300 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 22.0 (a) Determine the initial volume of the gas. b) Determine the temperature of the gas.

Explanation / Answer

Problem 1:

T1 =330 K , P = 2.5 kPa, V1 =1 m^3, V2 =3 m^3

Q = 12.2 kJ

In isobaric process Work done by gas W =PDV

W = (2.5)(3-1) = 5kJ

(a) From first law of thermodynamics

Q = DU +W

DU = Q-W = 12.2 -5

Change in internal energy DU = 7.2 kJ

(b) From ideal gas equation PV =nRT

For constant pressure

T2/T1 =V2/V1

T2 = (3/1)(330)

Final temperature T2 =990 K

Problem 2:

n =1 mol , W =3300 J

Pf = 1 atm =101325 Pa , Vf =22 L = 0.022 m^3

R =8.314 J/mol.K

From ideal gas equation PV =nRT

T = PV/nR = (101325*0.022)/(1*8.314)

T = 268.12 K

(a) WOrk done in isothermal process

W = nRTln (V2/V1)

3300 = (1*8.314*268.12) ln(0.022/V1)

Initial volme pf gas V1 = 0.005 m^3

(b) In isothermal process temperature is constant

T =268.12 K

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