Resources Set 13: Ch 18 HW Exercise 18.43 previous 9 of 12 next Exercise 18.43 P

Question

Resources Set 13: Ch 18 HW Exercise 18.43 previous 9 of 12 next Exercise 18.43 Part A Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol. C 741 J/(kg K) Submit My Answers Give up Correct Part B You warm 1.60 kg of water at a constant volume from 22.0 OC to 28.0 C in a kettle. For the same amount of heat, how many kilograms of 22.0 C C air would you be able to warm to 28.0 o C? Make the simplifying assumption that air is 100% N2 m 6.45 kg

Explanation / Answer

Part B

You have got Part A correct, so I will use value of specific heat capacity of N2, sn = 741 J/kg-K.

Heat needed for water to heat it from 220C to 280C is,

Qw = msw(28 - 22) = (1.6 kg)(4187 J/kg-K)(6) = 40195.2 J

If air containing 100% nitrogen is heated from 220C to 280C , heat required is,

Q = mnsn(28 - 22)

or, Qw = mnsn(28 - 22)

or, mnsn(28 - 22) = 40195.2 J

or, mn(741 J/kg-K)(28 - 22) = 40195.2 J

or, mn = 9 kg, is the required mass of nitrogen.

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Part C

Molar mass of nitrogen is 28 g/mol. So 0.028 kg of nitrogen is one mole of nitrogen,

So, 9 kg of nitrogen has number of moles n = 9 kg/0.028kg = 322.88 mol

So, using PV = nRT we get,

(101325 Pa)V = (322.88 mol) (8.31 J/mol-K)(273 + 22)K

or, V = 7.81 m3

or, V = 7.81X103 L is the required volume.

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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