1. Sababaraha gas idéal mimitina ngagaduhan tekenan P sareng volume V. Upami gas ngalaman prosés isotermal sahingga tekanan ahir janten 4 kali tekanan awal, maka volume gas ahir nyaéta…
Dipikanyaho:
Tekanan awal (P1) = P
Tekanan ahir (P2) = 4P
Volume awal (V1) = V
miharep: Volume ahir (V2)
Solusi:
Rumus tina Hukum Boyle :
PV = ajeg
P1 V1 = P2 V2
(P)(V) = (4P)(V)2)
V = 4 V2
V2 = V / 4 = ¼ V
Volume gas ahir nyaéta ¼ kali volume awal.
2. Dina wadah anu katutup, gas mekar sahingga final jilid janten 2 kali volume awal (V = volume awal, P = tekanan awal). Tekanan gas ahir nyaéta…
Dipikanyaho:
Tekanan awal (P1) = P
Volume awal (V1) = V
Volume ahir (V2) = 2V
hayang : Tekanan ahir (P2)
Solusi:
P1 V1 = P2 V2
PV = P2 (2V)
P= P2 (2)
P2 = P / 2 = ½ P
Tekanan gas janten ½ kali tekanan awal.
3. Dina wadah anu katutup, gas-gas anu tekananana 2 atm sareng volumena 1 liter. Upami tekanan gas janten 4 atm maka volume gas janten...
Dipikanyaho:
Tekanan awal (P1) = 2 atm = 2 x 105 Pa
Tekanan ahir (P2) = 4 atm = 4 x 105 Pa
Volume awal (V1) = 1 liter = 1 dm3 = 1 x 10-3 m3
hayang : Volume ahir (V2)
Solusi:
P1 V1 = P2 V2
(2 x 105)(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 = 0.5 x 10-3 m3 = 0.5 dm3 = 0.5 liters
