3 Tusaalooyin oo ka mid ah sharciga Boyle (heerkulka isothermal-constant)
1. Xaddiga gaaska ugu habboon marka hore wuxuu leeyahay cadaadis P ah iyo mug V ah. Haddii gaasku maro hab-socod isothermal ah si cadaadisku u noqdo 4 jeer cadaadiska asalka ah, mugga gaaska ayaa isu beddelaya...
Dood
Waa la garanayaa :
Cadaadiska bilowga ah (P)1) = P
Cadaadiska kama dambaysta ah (P)2) = 4P
Mugga bilowga ah (V)1) = V
La weydiiyay : mugga ugu dambeeya ee gaaska (V)2)
Jawab :
Sharciga Boyle (habka heerkulka isothermal ama heerkulka joogtada ah) :
PV = joogto ah
P1 V1 =P2 V2
(P)(V) = (4P)(V)2)
V = 4 V2
V2 = V / 4 = ¼ V
Mugga gaaska wuxuu isu beddelaa ¼ mugga bilowga ah.
2. Weel xiran, gaasku wuu fidaa si muggiisu u beddelo labanlaab mugga bilowga ah (V = mugga bilowga ah, P = cadaadiska bilowga ah). Cadaadiska gaasku wuxuu isu beddelaa…
Dood
Waa la garanayaa :
Cadaadiska bilowga ah (P)1) = P
Mugga bilowga ah (V)1) = V
Mugga kama dambaysta ah (V)2) = 2V
La weydiiyay : cadaadiska kama dambaysta ah (P)2)
Jawab :
P1 V1 =P2 V2
PV = P2 (2V)
P=P2 (2)
P2 = B / 2 = ½ B
Cadaadiska gaaska wuxuu isu beddelaa ½ jeer cadaadiska bilowga ah.
3. Weel xiran waxaa ku jira gaas oo leh cadaadis dhan 2 atm iyo mug dhan 1 litir. Haddii cadaadiska gaasku noqdo 4 atm markaas mugga gaasku wuxuu noqon doonaa...
Dood
Waa la garanayaa :
Cadaadiska bilowga ah (P)1) = 2 atm = 2 x 105 Pascal
Cadaadiska kama dambaysta ah (P)2) = 4 atm = 4 x 105 Pascal
Mugga bilowga ah (V)1) = 1 litir = 1 dm3 = 1x10-3 m3
La weydiiyay Mugga kama dambaysta ah (V)2)
Jawab :
P1 V1 =P2 V2
(2x105)(1 x 10-3) = (4 x 105) V2
(1)(1 x 10)-3) = (2) V2
1 x 10-3 = (2) V2
V2 = ½ x 10-3
V2 = 0,5x10-3 m3 = 0,5 dm3 = 0,5 litir
