User:JelloPiranha/Sandbox3

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

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

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 = ?''' $$(350.torr)(200.mL)=(700.torr)(x)$$ $$70000torrmL=700xtorr$$ $$100mL=x$$ $$V_2=100.mL$$

'''20b. P1 = 0.75 atm, V1 = ?, P2 = 0.48 atm, V2 = 435 mL''' $$(0.75atm)(x)=(0.48atm)(435mL)$$ $$0.75xatm=208.8atmmL$$ $$x=278.4mL$$ $$V_1=280mL$$

'''20c. P1 = ?, V1 = 2.4 × 105 L, P2 = 180 mm Hg, V2 = 1.8 × 103 L''' $$(24000L)(x)=(180mmHg)(1800L)$$ $$24000xL=324000mmHgL$$ $$x=13.5mmHg$$ $$P_1=14mmHg$$