i need help please

Question

11)A gas occupying 73 mL at standard conditions is heated to 16◦C while the pressure is reduced to 0.8152 atm. What is the new volume occupied by the gas? Answer in units of mL.

12)A sample of oxygen is collected in a 164 mL container over water at 15◦C, and the barometer reads 744 torr. What volume would the dry gas occupy at 768 torr and 15◦C? Water’s partial pressure at 15◦C is 12.8 torr. Answer in units of mL

Answer 1

Short answers: (11) 94.8 mL. (12) 156 mL.

11) Combined gas law. "Standard conditions" means 1.000 atm and 0 °C, and 0 °C is 273.15 K. Pressure, volume and temperature all change here while the amount of gas stays fixed, so this is the full combined gas law. Convert to kelvin first - using Celsius directly is the single most common way to lose this problem.

P1 V1 / T1 = P2 V2 / T2      so      V2 = V1 x (P1/P2) x (T2/T1)

V1 = 73 mL      P1 = 1.000 atm    T1 = 273.15 K
                P2 = 0.8152 atm   T2 = 16 + 273.15 = 289.15 K

V2 = 73 x (1.000/0.8152) x (289.15/273.15)
V2 = 73 x 1.2267 x 1.0586
V2 = 94.8 mL

Check it before you trust it. You heated the gas (volume up) and you lowered the pressure (volume up), so both changes push the same direction and the answer has to come out larger than 73 mL. It does. If your number had landed below 73 you would know immediately that one of the two ratios was flipped upside down - that sanity check catches the error without redoing the arithmetic. If your instructor wants the significant figures to match the 73 mL, report 95 mL.

12) Gas collected over water. When a gas is collected by bubbling it up into an inverted container of water, the space above the water holds two gases, not one: your oxygen, plus water vapor that evaporated into it. The 744 torr the barometer reads is the total pressure of that mixture. Dalton's law says the total is just the sum of the parts:

P(total) = P(O2) + P(water)
744 torr = P(O2) + 12.8 torr
P(O2)    = 744 - 12.8 = 731.2 torr

So 731.2 torr is what the oxygen alone is exerting, and "the dry gas" means exactly that oxygen with the water taken away. The temperature is 15 °C on both sides, so the combined law collapses - the T terms cancel and you are left with Boyle's law:

P1 V1 = P2 V2

V2 = V1 x P1/P2 = 164 mL x (731.2 torr / 768 torr)
V2 = 164 x 0.95208
V2 = 156 mL

The trap is using 744 torr instead of 731.2. That gives 159 mL - close enough to look perfectly reasonable and still be marked wrong, which is why this problem is assigned. Also note that the vapor pressure of water depends only on the temperature, never on how much gas you collected or how big the container is. That is why the problem can hand you 12.8 torr as a single number for 15 °C: it comes straight off a table.

One habit that makes both of these fast: before choosing a formula, write down what is being held constant. In 11 nothing is constant, so it is the full combined law. In 12 the temperature is constant, so two terms cancel and it is only Boyle. You never actually have to memorize Boyle, Charles and Gay-Lussac as three separate laws - each one is the combined law with something crossed out.