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Chemistry problem
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最佳解答:
by applying the general gas law (PV)/T = constant P1= 4.65 atm P2 = ? V1 = 38.0 L V2 = 55.0 L T1 = 35 + 273 = 308K T2 = 4 + 273 = 277K (4.65)(38.0)/ 308 = (P2)(55.0)/ 277 P2 = 2.89 atm
其他解答:
Ideal gas eq.: PV=nRT R is constant, n stays unchanged, P1 V1 / T1 = P2 V2 / T2 P2 = P1 V1 T2 / V2 T1...[1] where P1=4.65atm V1=38.0L T2=4degC=277K (always remember to use Kelvin) V2=55.0L T1=35degC=308K Sub. into [1] P2=(4.65atm)(38.0L)(277K) / (55.0L)(308K) P2=2.89atm Hence pressure in the tank is 2.89atm63D0B758E2D502CC
Chemistry problem
發問:
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2. A 38.0-L gas tank at 35celsius has nitrogen at a pressure of 4.65 atm. The contents of the tank are transferred without loss to an evacuated 55.0-L tank in a cold room where the temperature is 4celsius. What is the pressure in the tank?最佳解答:
by applying the general gas law (PV)/T = constant P1= 4.65 atm P2 = ? V1 = 38.0 L V2 = 55.0 L T1 = 35 + 273 = 308K T2 = 4 + 273 = 277K (4.65)(38.0)/ 308 = (P2)(55.0)/ 277 P2 = 2.89 atm
其他解答:
Ideal gas eq.: PV=nRT R is constant, n stays unchanged, P1 V1 / T1 = P2 V2 / T2 P2 = P1 V1 T2 / V2 T1...[1] where P1=4.65atm V1=38.0L T2=4degC=277K (always remember to use Kelvin) V2=55.0L T1=35degC=308K Sub. into [1] P2=(4.65atm)(38.0L)(277K) / (55.0L)(308K) P2=2.89atm Hence pressure in the tank is 2.89atm63D0B758E2D502CC
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