Question

1. what would be the new volume if 250 cm³ of gas at 25°c and 730 mm pressure were changed to standard conditions of temperature and pressure? (\\(\frac{n}{t}=constant\\))

initial p: 730 mm, t: 25 + 273 = 298 k, v: 250 cm³, n: -
final p: t: -
effect: -

1. sams bike tire contains 15 units of air particles and has a volume of 160ml. under these conditions the pressure reads 13 psi. the tire develops a leak. now it contains 10 units of air and has contracted to a volume of 150ml. what would the tire pressure be now?

initial p: 13 psi, t: -, v: 160ml, n: 15 u
final p:?, t: -, v: 150ml, n: 10 u
effect: -

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.

initial p: 93 kpa, t: -, v: 0.250l, n: 5.0 puffs
final p: -, t: -, v: 0.250l, n: 8.0 puffs
effect: -

1. 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

initial p: 1.20 atm, t: 303k, v: 350ml, n: -
final p: 1 atm, t:?, v: 250ml, n: -
effect: -

Explanation:

Step1: Identify the gas - law formula

For problems involving changes in pressure, volume, and temperature of a gas with constant amount of gas ($n$ is constant), we use the combined gas - law formula $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$.

Step2: Convert units if necessary

For problem 7:
Initial conditions: $P_1 = 1.20\ atm$, $T_1=(30 + 273)K=303K$, $V_1 = 350\ mL$
Final conditions: Standard pressure $P_2 = 1\ atm$, $V_2 = 250\ mL$, and we need to find $T_2$.

Step3: Rearrange the combined - gas law formula to solve for $T_2$

Starting from $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, we can cross - multiply to get $P_1V_1T_2=P_2V_2T_1$. Then $T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step4: Substitute the values into the formula

$T_2=\frac{1\ atm\times250\ mL\times303K}{1.20\ atm\times350\ mL}$
$T_2=\frac{250\times303}{1.20\times350}K$
$T_2=\frac{75750}{420}K\approx180.36K$
Converting back to Celsius, $t_2=(180.36 - 273)^{\circ}C\approx - 92.64^{\circ}C$

Answer:

For problem 7, the required temperature is approximately $180\ K$ or $-93^{\circ}C$ (rounded to the nearest whole number).

(Note: We can solve other problems in a similar fashion using appropriate gas - laws like Boyle's law ($P_1V_1 = P_2V_2$ for constant $T$ and $n$), Charles's law ($\frac{V_1}{T_1}=\frac{V_2}{T_2}$ for constant $P$ and $n$), or Avogadro's law ($\frac{V_1}{n_1}=\frac{V_2}{n_2}$ for constant $T$ and $P$) depending on which variables are constant in each case.)