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
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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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