Question

a gas is cooled from 365 k to 285 k while its volume changes from 12.8 l to 9.9 l. the initial pressure of the gas is 1.9 atm. what is the final pressure of the gas, rounded to the nearest tenth?
○ 1.3 atm
○ 2.9 atm
○ 1.9 atm
○ 9.9 atm

Explanation:

Step1: Recall Combined Gas Law

The combined gas law is $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, where $P_1, V_1, T_1$ are initial pressure, volume, temperature; $P_2, V_2, T_2$ are final ones.

Step2: Identify Given Values

$P_1 = 1.9\ \text{atm}$, $V_1 = 12.8\ \text{L}$, $T_1 = 365\ \text{K}$, $V_2 = 9.9\ \text{L}$, $T_2 = 285\ \text{K}$.

Step3: Solve for $P_2$

Rearrange the formula: $P_2=\frac{P_1V_1T_2}{V_2T_1}$.
Substitute values: $P_2=\frac{1.9\times12.8\times285}{9.9\times365}$.
Calculate numerator: $1.9\times12.8 = 24.32$; $24.32\times285 = 6931.2$.
Calculate denominator: $9.9\times365 = 3613.5$.
Divide: $P_2=\frac{6931.2}{3613.5}\approx1.92\approx1.9$? Wait, no, wait, miscalculation. Wait, $1.9\times12.8 = 24.32$; $24.32\times285$: $24\times285 = 6840$, $0.32\times285 = 91.2$, total $6840 + 91.2 = 6931.2$. Denominator: $9.9\times365$: $10\times365 = 3650$, minus $0.1\times365 = 36.5$, so $3650 - 36.5 = 3613.5$. Then $6931.2\div3613.5\approx1.92$, but wait, maybe I messed up. Wait, no, the combined gas law: when temperature decreases and volume decreases, pressure? Wait, initial T=365, final T=285 (decrease), initial V=12.8, final V=9.9 (decrease). So pressure: let's recalculate. Wait, $P_2=\frac{1.9\times12.8\times285}{9.9\times365}$. Let's compute step by step:

$1.9\times12.8 = 24.32$

$24.32\times285$: $24.32\times200 = 4864$, $24.32\times80 = 1945.6$, $24.32\times5 = 121.6$; total $4864 + 1945.6 = 6809.6 + 121.6 = 6931.2$

$9.9\times365$: $10\times365 = 3650$, minus $0.1\times365 = 36.5$, so $3650 - 36.5 = 3613.5$

Now, $6931.2\div3613.5\approx1.92$, but the options have 1.3, 2.9, 1.9, 9.9. Wait, maybe I used the wrong formula? Wait, no, combined gas law is $\frac{P1V1}{T1}=\frac{P2V2}{T2}$. So $P2 = P1\times\frac{V1}{V2}\times\frac{T2}{T1}$. So $\frac{V1}{V2}=\frac{12.8}{9.9}\approx1.2929$, $\frac{T2}{T1}=\frac{285}{365}\approx0.7808$. Then $P1\times1.2929\times0.7808 = 1.9\times(1.2929\times0.7808)\approx1.9\times1.009\approx1.917$, so approximately 1.9 atm. Wait, but let's check again. Wait, maybe the problem is that I misread the initial pressure? The initial pressure is 1.9 atm. So the calculation gives around 1.9 atm. So the answer is 1.9 atm.

Answer:

1.9 atm (the option: 1.9 atm)