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Problem 3. A box with dimensions of 30 cm x 50 cm x 10 cm is filled with argon at 15 °C and 1 atm of pressure (argon is a gas
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Given : V = 30 cm×50 cm× 10 cm ; T1 = 15°c ( 288 K); P1 = 1 atm = 1.013×105 Pa ; T2 = 20°c (293 K)

Constant : Molar mass of argon : M = 39.948 g/mol

Specific heat of argon : cp = 523 J/kgK

Universal gas constant : R = 8.31 J/mol-K

Solution:

Volume of Argon , V = 30×50×10 cm3 = 0.015 m3

Using ideal gas equation:

PV = nRT1

where n is the number of moles and R is the universal gas constant.

Putting values to get number of moles

(1.013×105Pa)(0.015 m3) = n(8.31 J/molK)(288 K)

n = 0.635 mol

Now , the mass of Argon is givem by:

m = (39.948 g/mol)×(0.635 mol) = 25.37 g

The heat added to raise the temperature is given by:

Q = mcp\DeltaT

= (25.37×10-3kg)(523 J/kgK)(293 K-288 K)

= 66.34 J

Answer : 66.34 J

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