A 1.531 sample of SO_2 at pressure of 5.6 Times 10^3 Pa. if the pressure changed
ID: 1052428 • Letter: A
Question
A 1.531 sample of SO_2 at pressure of 5.6 Times 10^3 Pa. if the pressure changed to 1.5 Times 10^2 Pa at constant temperature, what is the new volume of the gas. A gas sample with a volume of 2.58 I at 1.0 atm and 15 degree C. What volume this gas will occupy at 38 degree C and 1.0 atm? 12.2 I sample containing 0.5 mol oxygen at pressure of 1.0 atm and a temperature of 25 degree C. If all this O_2 was converted to ozone (O_3) at same temperature and pressure what is the volume of the ozone? A sample containing 0.614 moles of a gas at 12.0 degree C occupies a volume ot 12.9 I., What pressure does the gas exert? A gas at 34 degree C and 1.75 atm has a density of 3.4 g L Calculate the molar mass of the gas. A sample of methane (CH_2) at 0.848 atm and 4.0 degree C occupies a volume of 7.0 I., What volume will the gas occupy if the pressure increases to 1.52 atm and the temperature increased to 11.0 degree C. 0.5 1. of H_2 reacted with 0.6 I. of O_2 at STP according to the reaction: 2H_2 + O_2 rightarrow 2H_2O What volume water will occupy at 1.0 atm and 350 degree C? 46.0 I helium at 25 degree C and 1.0 atm was transferred with 12.0 I. O_2 at 25 degree C and 1.0 atm to a tank with a volume of 5.0 I., Calculate the partial pressure of each gas and the total pressure in the tank at 25 degree C. The partial pressure of O_2 was observed to be 156 torr in air with an atmospheric pressure of 743 torr. Calculate the mole fraction of O_2 present. A volume of 2.0 I of helium at 46 T and 1.2 atm was added to a vessel that contained 4.5 I. of N_2 at STP. What is the total pressure and partial pressure of each gas at 0 degree C alter the He was added?Explanation / Answer
We shall assume ideal behavior of gases for all the examples (since nothing really is mentioned).
1) Use Boyle’s law:
P1*V1 = P2*V2
where P1 = 5.6*103 Pa; P2 = 1.5*104 Pa; V1 = 1.53 L; V2 = ?
V2 = P1*V1/P2 = (5.6*103 Pa)*(1.53 L)/(1.5*104 Pa) = 0.5712 L (ans).
2) We note that the pressure is kept constant at 1.0 atm; we use Charles Law:
V1/T1 = V2/T2
We need to convert the given temperatures in C to K.
T1 = (15 + 273)K = 288 K
T2 = (38 + 273) K = 311 K
V1 = 2.58 L; V2 = ?
V2 = (V1/T1)*T2 = (2.58 L/288 K)*(311 K) = 2.786 L 2.79 L (ans).
3) 1.0 atm and 25C represent STP; therefore, we can use the concept of molar volume.
Molar volume is the volume occupied by 1 mole of a gas at STP.
Since 0.5 mole O2 occupies 12.2 L at STP, therefore, molar volume = (12.2 L/0.5 mol) = 24.4 L/mol.
Now, write down the conversion equation:
3 O2 (g) ------> 2 O3 (g)
The molar ratio of O2 and O3 is 3:2. Since we had 0.5 mol O2 at STP, moles of O3 produced = (0.5 mole O2)*(2 mole O3/3 mole O2) = 0.33 mole.
Volume occupied by 0.33 mole O3 = (24.4 L/mol)*(0.33 mole) = 8.052 L (ans).
4) Use the equation of state P*V = n*R*T where V = volume = 12.9 L; T = absolute temperature = (12.0 + 273) K = 285 K, n = moles of the gas = 0.614 mole and R = gas constant = 0.082 L-atm/mol.K.
Plug in the values to obtain
P = nRT/V = (0.614 mole)*(0.082 L-atm/mol.K)*(285 K)/(12.9 L) = 1.112 atm (ans).
5) Use the ideal gas law:
PV = nRT where n = m/M is the moles of the gas taken; m = mass of the gas taken and M = molar mass of the gas.
Again, d = m/V
Therefore,
P = (m/M)*RT/V = (m/V)*RT/M
Put d = m/V and write,
P = dRT/M
Given, d = 3.4 g/L; R = 0.082 L-atm/mol.K; T = (34 + 273) K = 307 K and P = 1.75 atm, we have,
1.75 atm = (3.4 g/L)*(0.082 L-atm/mol.K)*(307 K)/M
====> M = (3.4 g/L)*(0.082 L-atm/mol.K)*(307 K)/(1.75 atm) = 48.909 g/mol 48.91 g/mol.
The molar mass of the gas is 48.91 g/mol (ans).
6) Use the combined gas law:
P1V1/T1 = P2V2/T2
Here, T1 = (4.0 + 273) K = 277 K; T2 = (11.0 + 273) K = 284 K.
P1 = 0.848 atm; P2 = 1.52 atm and V1 = 7.0 L.
Therefore,
V2 = (P1V1/T1)*(T2/P2) = (0.848 atm)*(7.0 L)*(284 K)/(277 K).(1.52 atm) = 4.00 L (ans).
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