Question

part iii. intro to gasses
r = 0.08206 l - atm / mol - k
1.00 atm = 14.7 psi = 101.325 kpa = 760 torr = 760 mmhg = 29.92 inhg = 1.01325 bar

1. pressure conversions. convert 2.65 atm to:

bar
torr
psi
kpa

1. find the final volume for each gas sample.

a. a 2.65 l sample at 14.7 psi has a new pressure of 3.85 psi.
b. a 8.74 l sample at 28°c is heated to 150°c.
c. a 1.85 l sample at 15.0 psi and 25°c is now at 3.85 atm and -20°c.

Explanation:

Step1: Convert 2.65 atm to bar

Use the conversion factor 1 atm = 1.01325 bar.
$2.65\times1.01325 = 2.6851125$ bar

Step2: Convert 2.65 atm to torr

Use the conversion factor 1 atm = 760 torr.
$2.65\times760=2014$ torr

Step3: Convert 2.65 atm to psi

Use the conversion factor 1 atm = 14.7 psi.
$2.65\times14.7 = 38.955$ psi

Step4: Convert 2.65 atm to kPa

Use the conversion factor 1 atm = 101.325 kPa.
$2.65\times101.325=268.51125$ kPa

Step5: Solve part 5a

Use Boyle's law $P_1V_1 = P_2V_2$. Given $V_1 = 2.65$ L, $P_1=14.7$ psi, $P_2 = 3.85$ psi.
$V_2=\frac{P_1V_1}{P_2}=\frac{14.7\times2.65}{3.85}=\frac{38.955}{3.85}\approx10.12$ L

Step6: Solve part 5b

Use Charles's law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (temperatures in Kelvin). $T_1=28 + 273=301$ K, $T_2=150+ 273 = 423$ K, $V_1 = 8.74$ L.
$V_2=\frac{V_1T_2}{T_1}=\frac{8.74\times423}{301}=\frac{3707.02}{301}\approx12.32$ L

Step7: Solve part 5c

First convert pressures to same - unit (psi) and temperatures to Kelvin. 1 atm = 14.7 psi, so $P_2=3.85\times14.7 = 56.595$ psi. $T_1=25 + 273=298$ K, $T_2=-20 + 273 = 253$ K, $V_1 = 1.85$ L, $P_1 = 15.0$ psi.
Use the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$, then $V_2=\frac{P_1V_1T_2}{P_2T_1}=\frac{15.0\times1.85\times253}{56.595\times298}=\frac{15.0\times1.85\times253}{16865.31}\approx0.42$ L

Answer:

• 2.65 atm in bar: 2.6851125 bar
• 2.65 atm in torr: 2014 torr
• 2.65 atm in psi: 38.955 psi
• 2.65 atm in kPa: 268.51125 kPa
• 5a final volume: 10.12 L
• 5b final volume: 12.32 L
• 5c final volume: 0.42 L