Question

Page1 of 9 and 730 mmHg is cocled to give a pressure of 60 maHs ur answer to two significant figures and Include the appropriate unints, ANSWER Problem 8.36 A sample Part A e of argon gas has a volume ot 705 mL at a pressure of 1 20 atm and a remperature of 144 C What is the volume of the gas in millinters amount of gas remains the same? when the pressure and temperature of the ges sample are changed 1o 670 mmig o 670 mmHg, and 305 K ANSWER Part B ed to 0 950 atm and 82 C he What is the volume of the gas in mililiters when the pressure and temperature of the gas sample are chang amount of gas remains the same? ANSWER mL Part C What is the volume of the gas in mililiters when the amount of gas remains the same? i pressure and temperature of the gas sample are changed to 18 0 atm and-20 Ct the ANSWER mL A sample containing 1.50 moles of Ne gas has an initial volume of 8.00 L What is the final volume of the gas, in lters, when each of the to changes occurs in the quantity of the gas at constant pressure and temperature? Problem 8.41

Answer 1

Problem 8.36

Part A

$\frac{ P_1 \times V_1}{T_1}=\frac{ P_2 \times V_2}{T_2}$

676 mm Hg 760 mm Hg/atm × V, 1.20 atm × 705 mL (144 273.15) K 305 K

1.20 atrn × 705 mL 0.889 atm × V2 305 K 417.15 K

1.20 atm × 705 mL 305K 0.889 atm 417.15 K

Part B

$\frac{ P_1 \times V_1}{T_1}=\frac{ P_2 \times V_2}{T_2}$

$\frac{ 1.20 \ atm \times 705 \ mL}{(144+273.15) \ K}=\frac{ 0.950 \ atm \times V_2}{(82+273.15 ) \ K}$

$\frac{ 1.20 \ atm \times 705 \ mL}{417.15 \ K}=\frac{ 0.950 \ atm \times V_2}{355.15 \ K}$

$V_2=\frac{ 1.20 \ atm \times 705 \ mL}{417.15 \ K} \times \frac{355.15 \ K}{ 0.950 \ atm}$

Part C

$\frac{ P_1 \times V_1}{T_1}=\frac{ P_2 \times V_2}{T_2}$

$\frac{ 1.20 \ atm \times 705 \ mL}{(144+273.15) \ K}=\frac{ 18.0 \ atm \times V_2}{(273.15-20) \ K}$

$\frac{ 1.20 \ atm \times 705 \ mL}{ 417.15 \ K}=\frac{ 18.0 \ atm \times V_2}{ 253.15 \ K}$

$V_2=\frac{ 1.20 \ atm \times 705 \ mL}{ 417.15 \ K} \times \frac{253.15 \ K}{18.0 \ atm }$