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
TI 10 Lof a gas are found to exert 0.81914631 atm at 35.0°C. What is the required temprerature to change the pressu...
10 L of a gas are found to exert 0.81914631 atm at 35.0°C. What is the required temperature to change the pressure to standard pressure? Give your answer in Celsius (assume constant volume) (10 pts)
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