1 atm is how many mmhg

1 atm is how many mmhg

Ideal Gas Law: Problem Set #1

Problem #1: Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.

Solution:

1) Rearrange PV = nRT to this: V = nRT / P

2) Substitute: V = [ (2.34 g / 44.0 g mol¯ 1 ) (0.08206 L atm mol¯ 1 K¯ 1 ) (273.0 K) ] / 1.00 atm

V = 1.19 L (to three significant figures)

Problem #2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample.

Solution:

1) Rearrange PV = nRT to this: n = PV / RT

2) Substitute: n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯ 1 K¯ 1 ) (273.0 K) ]

n = 2.50866 mol (I'll keep a few guard digits)

3) Multiply the moles by the atomic weight of Ar to get the grams: 2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)

Problem #3: At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

Solution:

1) Rearrange PV = nRT to this: T = PV / nR

2) Substitute: T = [ (1.95 atm) (12.30 L) ] / [ (0.654 mol) (0.08206 L atm mol¯ 1 K¯ 1 ) ]

T = 447 K

Problem #4: A 30.6 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?

Solution: Since one mole of gas occupies 22.414 L at STP, the molecular weight of the gas is 30.6 g mol¯ 1

Problem #5: A 40.0 g gas sample occupies 11.2 L at STP. Find the molecular weight of this gas.

Solution: 11.2 L at STP is one-half molar volume, so there is 0.500 mol of gas present. Therefore, the molecular weight is 80.0 g mol¯ 1

Problem #6: A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?

Solution:

This problem, as well as the two just above can be solved with PV = nRT. You would solve for n, the number of moles. Then you would divide the grams given by the mole calculated.

1) Use PV = nRT: (1.00 atm) (19.2 L) = (n) (0.08206) (273 K)

n = 0.8570518 mol (I'll keep a few guard digits)

2) Determine the molecular weight: 12.0 g / 0.8570518 mol = 14.0 g/mol

3) Since it is at STP, we can also use molar volume: (19.2 L / 12.0 g) = (22.414 L / x )

19.2x = 268.968

x = 14.0 g/mol

Warning: you can only use molar volume when you are at STP.

Problem #7: 96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

Solution:

1) Solve for the moles using PV = nRT: n = PV / RT

n = [ (700.0 mmHg / 760.0 mmHg atm¯ 1 ) (48.0 L) ] / [ (0.08206 L atm mol¯ 1 K¯ 1 ) (293.0 K) ]

n = 1.8388 mol

2) Divide the grams given (96.0) by the moles just calculated above: 96.0 g / 1.8388 mol = 52.2 g/mol

Problem #8: 20.83 g of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

Solution:

1) Solve for the moles using PV = nRT: n = PV / RT

n = [ (79.97 kPa / 101.325 kPa atm¯ 1 ) (4.167 L) ] / [ (0.08206 L atm mol¯ 1 K¯ 1 ) (303.0 K) ]

n = 0.13227 mol

2) Divide the grams given (20.83) by the moles just calculated above: 20.83 g / 0.13227 mol = 157.5 g/mol

Notice that, in the two problems just above, the I converted the pressure unit given in the problem to atmospheres. I did this to use the value for R that I have memorized. There are many different ways to express R, it's just that L-atm/mol-K is the unit I prefer to use, whenever possible.

Also, you cannot use molar volume since the two problems just above are not at STP.

Problem #9: What is the value of and units on R? What is R called ("A letter" is not the correct answer!)? R is called the gas constant. It was first discovered, as part of the discovery in the mid-1830's by Emil Clapeyron of what is now called the Ideal Gas Law.

Sometimes it is called the universal constant because it shows up in many non-gas-related situations. However, it is mostly called the gas constant.

Depending on the units selected, the "value" for R can take on many different forms. Here is a list. Keep in mind these different "values" represent the same thing.

Problem #10: 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?

Solution:

1) Determine moles of CO2. 5.600 g / 44.009 g/mol = 0.1272467 mol

2) Use PV = nRT (P) (4.00 L) = (0.1272467 mol) (0.08206) (300 K)

P = 0.7831 atm (to four sig figs)

Source: www.chemteam.info

Category: Bank

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