Question

1. a closed flask of air (0.250l) contains 5.0 “puffs” of particles. the pressure probe on the flask reads 93 kpa. a student uses a syringe to add an additional 3.0 “puffs” of air through the stopper. find the new pressure inside the flask.
2. a 350 ml sample of gas has a temperature of 30°c and a pressure of 1.20 atm. what temperature would needed for the same amount of gas to fit into a 250 ml flask at standard pressure? 30 + 273 = 303k

Explanation:

Step1: Identify the relevant gas - law

For the first problem, since volume and temperature are constant (closed flask, no mention of temperature change), we use Avogadro's law in the form of $P_1/n_1 = P_2/n_2$ (derived from the ideal gas law $PV = nRT$). Initial number of moles $n_1$ is proportional to 5.0 puffs, and final number of moles $n_2$ is proportional to $5.0 + 3.0=8.0$ puffs. Initial pressure $P_1 = 93$ kPa.

Step2: Calculate the new pressure

We have $\frac{P_1}{n_1}=\frac{P_2}{n_2}$, so $P_2=\frac{n_2}{n_1}\times P_1$. Substituting $n_1 = 5.0$, $n_2 = 8.0$ and $P_1 = 93$ kPa, we get $P_2=\frac{8.0}{5.0}\times93$ kPa.
$P_2 = 148.8\approx149$ kPa.

For the second problem, we use the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. Standard pressure $P_2 = 1$ atm, $P_1 = 1.20$ atm, $V_1 = 350$ mL, $V_2 = 250$ mL, $T_1=(30 + 273)$ K$=303$ K.

Step3: Rearrange the combined - gas law for $T_2$

$T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step4: Substitute the values

$T_2=\frac{1\ atm\times250\ mL\times303\ K}{1.20\ atm\times350\ mL}$.
$T_2=\frac{250\times303}{1.20\times350}$ K.
$T_2=\frac{75750}{420}$ K$\approx180.36$ K. Converting to Celsius, $t_2=T_2 - 273=180.36 - 273=- 92.64\approx - 92.6$°C.

Answer:

1. The new pressure in the first flask is approximately 149 kPa.
2. The required temperature in the second case is approximately - 92.6°C.