1. Ka whakatakotoria e au ngā hau ki roto i tētahi ipu kua kati , ā, i te tīmatanga, ko te rōrahi he V, ko te pāmahana he T. Ko te pāmahana whakamutunga he 5/4T, ā, ko te pēhanga whakamutunga he 2P. He aha te rōrahi whakamutunga o te hau?
Mōhiotia:
Te rōrahi tīmatanga (V 1 ) = V
Te pāmahana tīmatanga (T 1 ) = T
Te pāmahana whakamutunga (T 2 ) = 5/4 T
Pēhanga tīmatanga (P 1 ) = P
Pēhanga whakamutunga (P 2 ) = 2P
E hiahiatia ana: Te Pukapuka Whakamutunga (V 2 )
Rongoā:
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2. Tātaihia te rōrahi o te 2.00 mol o ngā hau (hau pai) i te STP. STP = Te Pāmahana me te Pēhanga Paerewa.
Mōhiotia:
Ngā ira hau (n) = 2 ngā ira hukapapa
Pāmahana paerewa (T) = 0 ° C = 0 + 273 = 273 Kelvin
Pēhanga paerewa (P) = 1 atm = 1.013 x 10 5 Pa
Taarua hau ao (R) = 8.315 Joule/mol e .Kelvin
E hiahiatia ana : Te rōrahi o ngā hau (V)
Rongoā:
Te whārite o te ture hau pai (i roto i te maha o ngā ira, n)
Ko te rōrahi o ngā mole hau e 2 he 44.8 rita.
Ko te rōrahi o te 1 mole o ngā hau he 45.4 rita / 2 = 22.4 rita.
Ko te rōrahi o te 1 mole o tētahi hau he 22.4 rita.
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3. E 27°C te pāmahana o te hau hāora e 4 rita, ā, ko te pēhanga he 2 atm (1 atm = 105 Pa ) i roto i tētahi ipu kua kati. Ko te pūmau hau whānui (R) = 8.314 J.mol e −1 .K −1 me te tau a Avogadro ( N A ) = 6.02 x 1023 ngota/mol e . He aha ngā ngota o ngā hau hāora i roto i te ipu?
Mōhiotia:
Te rōrahi o ngā hau (V) = 4 rita = 4 dm³ = 4 x 10 -3 m³
Te pāmahana o ngā hau (T) = 27 ° C = 27 + 273 = 300 Kelvin
Te pēhanga o ngā hau (P) = 2 atm = 2 x 10 5 Pa
Pūmau hau whānui (R) = 8.314 J.mol e −1 .K −1
Te tau o Avogadro (N A ) = 6.02 x 10 23
E hiahiatia ana : He aha ngā ngota o ngā hau hāora i roto i te ipu (N)
Rongoā:
I roto i te 1 mole o ngā hau hāora, e 1.93 x 1023 ngā ngota hāora.
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4. Ko te rōrahi o tētahi ipu kei roto he hau neon (Ne, papatipu ngota = 20 u) i te pāmahana me te pēhanga paerewa (STP ) he 2 m³ . Tātaihia te papatipu o te hau neon!
Mōhiotia:
Papatipu ngota o te neon = 20 karamu/mol e = 0,02 kg/mol e
Pāmahana paerewa (T) = 0 ° C = 273 Kelvin
Pēhanga paerewa (P) = 1 atm = 1.013 x 10 5 Pascal
Rōrahi (V) = 2 m3
E hiahiatia ana : te taumaha (m) o te hau neon
Rongoā:
I te pāmahana me te pēhanga paerewa (STP), 1 mole o ngā hau katoa, whakaurua te hau neon, he rōrahi 22.4 ritas = 22.4 dm3 = 0.0448 m3.
I roto i te rōrahi o te 2 m 3 , e 44.6 ngā mole o te hau neon.
Ko te papatipu ngota whanaunga o te hau neon he 20 karamu/mole.
Ko te tikanga o tēnei, i roto i te 1 ira he 20 karamu, arā, 0.02 kg o te hau neon. Nā te mea he 0.02 kg o te hau neon i roto i te 1 ira, nō reira, i roto i te 44.6 ira he 44.6 ira x 0.02 kg/ira = 0.892 kg = 892 karamu o ngā hau neon.
