Answer:
Q1: 2.068 atm.
Q2: 0.7346 atm.
Q3: 336.7 mm Hg.
Q4: 393.7 K = 120.7 °C.
Q5: 694.1 mm Hg.
Q6: 397.1 K.
Q7: 676.5 mm Hg.
Q8: 72.68 kPa.
Explanation:
- <em>Gay-Lussac's law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas (K), when the volume is kept constant.</em>
- It can be written as: <em>P/T = constant.</em>
- So, for two different T and P we can express the relation as:
<em>P₁T₂ = P₂T₁.</em>
<em></em>
<em>Q1: </em>
P₁ = 2.0 atm, T₁ = 20.0°C + 273 = 293 K.
P₂ = ??? atm, T₂ = 30.0°C + 273 = 303 K.
∴ P₂ = P₁T₂/T₁ = (2.0 atm)(303 K)/(293 K) = 2.068 atm.
<em>Q2: </em>
P₁ = 0.55 atm, T₁ = 25.0°C + 273 = 298 K.
P₂ = ??? atm, T₂ = 125.0°C + 273 = 398 K.
∴ P₂ = P₁T₂/T₁ = (0.55 atm)(398 K)/(298 K) = 0.7346 atm.
<em>Q3: </em>
P₁ = 87.0 mm Hg, T₁ = 77.0 K.
P₂ = ??? atm, T₂ = 25.0°C + 273 = 298.0 K.
∴ P₂ = P₁T₂/T₁ = (87.0 mm Hg)(298 K)/(77 K) = 336.7 mm Hg.
<em>Q4: </em>
P₁ = 248.0 kPa, T₁ = 10.0°C + 273 = 283 K.
P₂ = 345.0 kPa, T₂ = ??? K.
∴ T₂ = P₂T₁/P₁ = (345.0 kPa)(283 K)/(248.0 kPa) = 393.7 K - 273 = 120.7 °C.
<em>Q5: </em>
P₁ = 750.0 mm Hg, T₁ = 22.0°C + 273 = 295 K.
P₂ = ??? mm Hg, T₂ = 0.0°C + 273 = 273 K.
∴ P₂ = P₁T₂/T₁ = (750.0 mm Hg)(273 K)/(295 K) = 694.1 mm Hg.
<em>Q6: </em>
P₁ = 599.0 mm Hg, T₁ = 40.0°C + 273 = 313 K.
P₂ = 760.0 mm Hg, T₂ = ??? K.
∴ T₂ = P₂T₁/P₁ = (760.0 mm Hg)(313 K)/(599 mm Hg) = 397.1 K.
<em>Q7: </em>
P₁ = 800.0 mm Hg, T₁ = 323 K.
P₂ = ??? mm Hg, T₂ = 273.15 K.
∴ P₂ = P₁T₂/T₁ = (800.0 mm Hg)(273.15 K)/(323 K) = 676.5 mm Hg.
<em>Q8: </em>
P₁ = 30.0 kPa, T₁ = - 150.0°C + 273 = 123 K.
P₂ = ??? kPa, T₂ = 25.0°C + 273 = 298 K.
∴ P₂ = P₁T₂/T₁ = (30.0 kPa)(298 K)/(123 K) = 72.68 kPa.
