User:JelloPiranha/Sandbox3
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[edit] 18.
Convert each of the following Celsius temperatures to Kelvin temperatures.
18a. 0.°C
0.°C + 273.15 = 273.15 K → 273 K
18b. 27°C
27°C + 273.15 = 300.15 K → 300 K
18c. -50.°C
-50.°C + 273.15 = 223.15 K → 223 K
18d. -273°C
-273°C + 273.15 = 0.15 K → 0 K
[edit] 19.
Convert each of the following Kelvin temperatures to Celsius temperatures.
19a. 273 K
273 K - 273.15 = -0.15°C → 0°C
19b. 350 K
350 K - 273.15 = 76.85°C → 77°C
19c. 100. K
100. K - 273.15 = -173.15°C → -173°C
19d. 20. K
20. K - 273.15 = -253.15°C → -253°C
[edit] 20.
Use Boyle's law to solve for the missing value in each of the following.
20a. P1 = 350. torr, V1 = 200. mL, P2 = 700. torr, V2 = ?
(350.torr)(200.mL) = (700.torr)(x)
70000torrmL = 700xtorr
100mL = x
V2 = 100.mL
20b. P1 = 0.75 atm, V1 = ?, P2 = 0.48 atm, V2 = 435 mL
(0.75atm)(x) = (0.48atm)(435mL)
0.75xatm = 208.8atmmL
x = 278.4mL
V1 = 280mL
20c. P1 = ?, V1 = 2.4 × 105 L, P2 = 180 mm Hg, V2 = 1.8 × 103 L
(24000L)(x) = (180mmHg)(1800L)
24000xL = 324000mmHgL
x = 13.5mmHg
P1 = 14mmHg
