Question

TI 10 Lof a gas are found to exert 0.81914631 atm at 35.0°C. What is the required temprerature to change the pressure to standard pressure? Give your answer in Celsius (assume constant volume) (10 pts) 4.73 L of a gas is collected at 32.0°C and 625.0 mmHg. When the gas is changed to STP conditions, what is the new volume? (10 pts) A commercial airliner has an internal pressure of 1.00 atm and temperature of 25° C at takeoff If the temperature of the airliner drops to 17C during the flight, what is the new cabin pressure? (assume constant volume) (10 pts) n0(8A)

Answer 1

Q4. According to Gay-Lussac's law, absolute temperature of an ideal gas is directly proportional to its pressure provided volume and amount of gas remains constant.

Mathematically, T1 / P1 = T2 / P2

where T1 = initial temperature = 35.0 ^oC = (35.0 + 273) K = 308 K

P1 = 0.81914631 atm

T2 = final temperature

P2 = final pressure = standard pressure = 1 atm

Substituting the values.

(308 K) / (0.81914631 atm) = T2 / (1 atm)

T2 = (308 K) * (1 atm / 0.81914631 atm)

T2 = (308 K) * (1.22)

T2 = 376 K

T2 = (376 - 273) K

T2 = 103.0 ^oC

Required temperature is 103.0 ^oC

Q5. According to combined gas law

(P1 * V1) / T1 = (P2 * V2) / T2

where P1 = initial pressure = 625.0 mmHg

V1 = initial volume = 4.73 L

T1 = initial temperature = 32.0 ^oC = (32.0 + 273) K = 305 K

P2 = final pressure = pressure at STP = 1 atm = 760 mmHg

V2 = final volume

T2 = final temperature = temperature at STP = 273 K

Substituting the values,

(625.0 mmHg * 4.73 L) / (305 K) = (760 mmHg * V2) / (273 K)

V2 = (4.73 L) * (625.0 mmHg / 760 mmHg) * (273 K / 305 K)

V2 = (4.73 L) * (0.82237) * (0.895)

V2 = 3.48 L

new volume is 3.48 L

Q6. According to Gay-Lussac's law, absolute temperature of an ideal gas is directly proportional to its pressure provided volume and amount of gas remains constant.

Mathematically, T1 / P1 = T2 / P2

where T1 = initial temperature = 25.0 ^oC = (25.0 + 273) K = 298 K

P1 = 1.00 atm

T2 = final temperature = 17.0 ^oC = (17.0 + 273) K = 290 K

P2 = final pressure

Substituting the values.

(298 K) / (1.00 atm) = (290 K) / (P2)

P2 = (1.00 atm) * (290 K / 298 )

P2 = (1.00 atm) * (0.973)

P2 = 0.973 atm

new cabin pressure = 0.973 atm