Chemistry, asked by mpmpmpmmp, 5 months ago

1.At what temperature will 41.6 g of N2(g) exert a pressure of 108.6 kPa in a 20.0L cylinder?

Question 7 options:

134 K


176 K


238 K


337 K

Answers

Answered by via77
0

Answer:

176 K

Explanation:

PV = n RT

1bar = 100000 pa

1 k Pa = 1000Pa

therefore

1 08.6 K Pa = 108.6 / 100 bar

28 g N2 = 28 g ( acc to mole concept )

therefore 41.6 g of N2 = 41.6 ÷ 28 = 1.48 moles of N2

108.6 / 100 x 20 L = 1.48 moles × 8.314 x T

T = 176 K

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