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A monatomic ideal gas in a rigid container is heated from 221 K to 285 K by adding 8130 J of heat. How many moles of gas are there in the container?

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princessesther2011

Answer:

15.28 moles

Explanation:

E = nR(T2 - T1)

E is the heat added = 8130 J

R is the gas constant = 8.314 J/mol.K

T2 is the final temperature of the gas = 285 K

T1 is the initial temperature of the gas = 221 K

8130 = n×8.314(285 - 221)

n = 8130/8.314×64 = 8130/532.096 = 15.28 moles

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