Homework 3 Problem 1 For each of the following gas law problems; (1) state the i
ID: 1028370 • Letter: H
Question
Homework 3 Problem 1 For each of the following gas law problems; (1) state the information given in the problem, (2) state what the problem is asking for, (3) specify whieh law (or equation) relates all of the variables involved int the problem, and (4) solve the problem. a. A syringe containing 8.76 mL of carbon dioxide gas initially at 25 C is placed into an ice bath. If the volume of the gas decreases to 2.92 mL what is the final temperature ('C) of the gas in the syringe assuming the pressure is held constant? b. A 2.50-L container holds nitrogen gas initially at a pressure of 1.75 atm. The gas is then allowed to expand, at a constant temperature, until it reaches a pressure of 1.00 atm. What is the new volume of the nitrogen gas? c. A 355-mL vessel contains 0.500 moles of water vapor at 112°C. What d. The density of a particular gas at STP is 3.16g/L. What is the molar e. A rigid container is filled with 173 Torr of air at 298K. The air inside is the pressure (in atm) of the water vapor in the vessel? mass of the gas? the container is then heated to 453 Torr. What is the pressure of the air at this temperature? f. A cylinder with a movable piston contains 0.233 moles of ammouia gas (NHs) and has a volume of 315 mL. What is the volume after an additional 0.254 moles of ammonia is added to cylinder while the pressure and temperature is held constant?Explanation / Answer
a) The initial volume of the carbon dioxide gas is V1 = 8.76 mL at temperature T1 = 25°C = (25 + 273) K = 298 K. The final volume of the gas is V2 = 2.92 mL.
The pressure and the mass of the gas are held constant.
We are required to find out the final temperature T2 of the gas.
Use Charle’s law as
V1/T1 = V2/T2 where T2 = final temperature of the gas.
=====> T2 = V2/V1*T1 = (2.92 mL/8.76 mL)*(298 K) = 99.33 K = (99.33 – 273)°C = -173.67°C (ans).
b) The initial volume of the nitrogen gas is V1 = 2.50 L at pressure P1 = 1.75 atm. The final pressure of the gas is P2 = 1.00 atm.
The temperature and the mass of the gas are held constant.
We are required to find out the final volume V2 of the gas.
Use Boyle’s law as
P1*V1 = P2*V2
=====> (1.75 atm)*(2.50 L) = (1.00 atm)*V2
=====> V2 = (1.75 atm)*(2.50 L)/(1.00 atm) = 4.375 L (ans).
c) The volume of the water vapor in the vessel is V = 355 mL = (355 mL)*(1 L/1000 mL) = 0.355 L. The temperature of the water vapor in the vessel is T = 112°C = (112 + 273) K = 385 K while the number of moles of water vapor in the vessel is n = 0.500 mole.
We are required to find out the pressure P of the gas.
Use the ideal gas law as
P*V = n*R*T
=====> P*(0.355 L) = (0.500 mole)*(0.082 L-atm/mol.K)*(385 K)
=====> P*(0.355 L) = 15.785 L-atm
=====> P = (15.785 L-atm)/(0.355 L) = 44.4648 atm 44.46 atm (ans).
d) The density of the gas is d = 3.16 g/L at STP [standard temperature and pressure; standard temperature T = 25°C = (25 + 273) K = 298 K and standard pressure P = 1.00 atm].
Let us assume that we took m g of the gas; therefore,
d = m/V
=====> V = m/d where V is the volume occupied by the gas.
Plug in the ideal gas equation as
P*V = n*R*T
====> P*(m/d) = n*R*T
Again, n = m/M is the number of moles of the gas taken and M is the molar mass of the gas. Therefore,
P*(m/d) = (m/M)*R*T
=====> P/d = RT/M
=====> M = d*R*T/P
=====> M = (3.16 g/L)*(0.082 L-atm/mol.K)*(298 K)/(1.00 atm) = 77.21776 g/mol 77.22 g/mol.
The molar mass of the gas is 77.22 g/mol (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.