Question

1. a 350 ml sample of gas has a temperature of 30°c and a pressure of 1.20 atm. what temperature would be needed for the same amount of gas to fit into a 250 ml flask at standard pressure? 30 + 273 = 303k

	p	t	v	n
final			250ml	--
effect			↓	--

Explanation:

Step1: Identify the combined - gas law

The combined - gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, where $P$ is pressure, $V$ is volume, and $T$ is temperature. Standard pressure $P_2 = 1\ atm$. Initial pressure $P_1=1.20\ atm$, initial volume $V_1 = 350\ mL$, initial temperature $T_1=30 + 273=303\ K$, and final volume $V_2 = 250\ mL$.

Step2: Rearrange the combined - gas law for $T_2$

From $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, we can get $T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step3: Substitute the values

Substitute $P_1 = 1.20\ atm$, $V_1=350\ mL$, $T_1 = 303\ K$, $P_2 = 1\ atm$, and $V_2 = 250\ mL$ into the formula:
$T_2=\frac{1\ atm\times250\ mL\times303\ K}{1.20\ atm\times350\ mL}$.
First, calculate the numerator: $1\times250\times303 = 75750$.
Then, calculate the denominator: $1.20\times350=420$.
$T_2=\frac{75750}{420}\ K\approx180.36\ K$.
Convert to Celsius: $t=T - 273=180.36 - 273=- 92.64^{\circ}C\approx - 92.6^{\circ}C$.

Answer:

$-92.6^{\circ}C$