Question

11 formula 1 punto determine the pressure, in atm, when a constant volume of gas at 1.2 atm is heated from 29.8 °c to 169.9 °c respuesta 12 formula 1 punto convert 8.2 atm to psi. (1 atm = 14.6 psi) round to 1 decimal place. respuesta 119.7 13 formula 1 punto a gas balloon has a volume of 2.6 liters when the temperature is 11.2 °c and the pressure is 743.1 mm of mercury. what will its volume be in l at 21.9 °c and 720.2 mm of mercury pressure? express your answer with 1 decimal place. respuesta

Explanation:

Step1: Identify the relevant gas - law for question 11

For a constant - volume process, we use Gay - Lussac's law: $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, where $P$ is pressure and $T$ is temperature in Kelvin. First, convert the Celsius temperatures to Kelvin.
$T_1 = 29.8+273.15=302.95$ K, $T_2 = 169.9 + 273.15=443.05$ K, and $P_1 = 1.2$ atm.

Step2: Solve for $P_2$

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can re - arrange to get $P_2=\frac{P_1T_2}{T_1}$.
Substitute the values: $P_2=\frac{1.2\times443.05}{302.95}\approx1.7$ atm.

Step3: Solve question 12

We are given the conversion factor $1$ atm $ = 14.6$ psi. To convert $8.2$ atm to psi, we use the formula $P_{psi}=P_{atm}\times14.6$.
$P_{psi}=8.2\times14.6 = 119.7$ psi.

Step4: Identify the relevant gas - law for question 13

We use the combined gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. First, convert the Celsius temperatures to Kelvin: $T_1=11.2 + 273.15 = 284.35$ K, $T_2=21.9+273.15 = 295.05$ K. Also, convert the pressures: $P_1 = 743.1$ mmHg, $P_2 = 720.2$ mmHg, and $V_1 = 2.6$ L.
Re - arrange the combined gas law for $V_2$: $V_2=\frac{P_1V_1T_2}{T_1P_2}$.
Substitute the values: $V_2=\frac{743.1\times2.6\times295.05}{284.35\times720.2}\approx2.8$ L.

Answer:

1. $1.7$ atm
2. $119.7$ psi
3. $2.8$ L