# Ideal Gas Laws with Temperature, Pressure and Volume

What is the temperature of 0.52mol of gas at a pressure of 1.3atm and a volume of 11.7L ?

This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm , 20.0 ∘ C , and 1.00 L . This information will apply to all parts of this problem A, B, and C.What will the pressure inside the container become if the piston is moved to the 2.00 L mark while the temperature of the gas is kept constant?

A weather balloon is inflated to a volume of 29.0L at a pressure of 744mmHg and a temperature of 31.3 ∘ C . The balloon rises in the atmosphere to an altitude, where the pressure is 360mmHg and the temperature is -14.1 ∘ C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude

Calculate the density of oxygen, O 2 , under each of the following conditions:•STP

•1.00 atm and 25.0 ∘ C

To identify a diatomic gas (X 2 ), a researcher carried out the following experiment: She weighed an empty 1.0-L bulb, then filled it with the gas at 1.30atm and 26.0 ∘ C and weighed it again. The difference in mass was 1.5g . Identify the gas.

A 1-L flask is filled with 1.25g of argon at 25 ∘ C . A sample of ethane vapor is added to the same flask until the total pressure is 1.05atm .What is the partial pressure of argon, P Ar , in the flask?What is the partial pressure of ethane, P ethane , in the flask?

Imagine that you have a 7.00L gas tank and a 3.00L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 125atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.

13.0 moles of gas are in a 3.00L tank at 20.9 ∘ C . Calculate the difference in pressure between methane and an ideal gas under these conditions. The van der Waals constants for methane are a=2.300L 2 ⋅atm/mol 2 and b=0.0430 L/mol .

To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 4.00mol of gas D measured at 251 ∘ C in a 1.75-L container assuming ideal behavior?

To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to small volumes. What is the pressure of 4.00mol of gas D measured at 251 ∘ C in a 1.75-L container assuming real behavior?

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#### Solution Preview

Gases

What is the temperature of 0.52mol of gas at a pressure of 1.3atm and a volume of 11.7L ?

Use the Ideal Gas Law :

PV = nRT

P = Pressure of confined gas in atmospheres

V= Volume of the confined gas, in liters

N= Number of moles of gas

R= Gas Constant, 0.0821 L atm/ mol K

T = Temperature in Kelvin

So to find the temperature =

T = PV/nR

= 1.3 atm * 11.7 L/ 0.52 mol * 0.0821 L *atm/mol*K

= 357.04 K

This figure (Figure 1) shows a container that is sealed at the top by a movable piston. Inside the container is an ideal gas at 1.00 atm , 20.0 ∘ C , and 1.00 L . This information will apply to all parts of this problem A, B, and C.What will the pressure inside the container become if the piston is moved to the 2.00 L mark while the temperature of the gas is kept constant?

Use Ideal Gas Law

P1 = 1.00 atm P2= ?

V1 = 1.00L V2 = 2.00 L

T K = Tc + 273.15

= 20° C + 273.15

= 293.15 K

PV = nRT

Set the P1V1 = P2V2, so

P2 = P1V1/V2

= 1 atm * 1L/ 2 L

= 0.5 atm

A weather balloon is inflated to a volume of 29.0L at a pressure of 744mmHg and a temperature of 31.3 ∘ C . The balloon rises in the atmosphere to an altitude, where the pressure is 360mmHg and the temperature is -14.1 ∘ C.Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude

Use Ideal gas law

V1 = 29.0 L

P1 = 744 mmHg

T = 31.3 C

Tk1 = Tc1 + 273.15

= 31.3 C + 273.15

= 304.45 K

P2 = 360 mmHg

Tk2 = Tc2 + 273.15

= -14.1 C + 273.15

= 259.05 K

V2= ?

P1V1 = nRT1

P2V2 = nRT2

P1V1/T1 = nR

P2V2/T2 = nR

P1V1/T1 = P2V2/T2

V2 = P1V1T2/ T1P2

V2 = 744 mmHg 29.0L 259.05K/304.45 K 360 mmHg

= 50.99 L

Calculate the density of ...

#### Solution Summary

The ideal gas law relates the four variables of pressure, volume, temperature, and number of moles of gas within a closed system. The ideal gas law takes the form of PV= nRT. The ideal gas law is the combination of all the simple gas laws such as Boyle's Law, Charles' Law, and Avogadro's Law. These simple gas law can be derived from the ideal gas law equation. The ideal gas law have to assume that no forces acting among the particles of gases and that the particles do not take up any space. The ideal gas law equation can be use to solve a gas problem when amount of gas is given and the mass of the gas is constant. Below are examples of how the ideal gas equation is used to solve gas problems.

Experiment to demonstrate the Ideal Gas Law.

Experiment to demonstrate the Ideal Gas Law.

Attached is the directions and data sheet. Here is the link for the data sheet. http://www.mhhe.com/physsci/physical/giambattista/thermo/thermodynamics.html

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Data and related questions

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Part 1

P (Pa) V (m^3) T(K)

498.6 5 300

453.27 5.5 300

415.5 6 300

383.53 6.5 300

356.14 7 300

332.4 7.5 300

311.62 8 300

293.29 8.5 300277 9 300

262.42 9.5 300

249.3 10 300

P (Pa) V (m^3) T (K)

249.3 10 300

262.42 9.5 300

277 9 300

293.29 8.5 300

311.62 8 300

332.4 7.5 300

356.14 7 300

383.53 6.5 300

415.5 6 300

453.27 5.5 300

498.6 5 300

554 4.5 300

623.25 4 300

712.28 3.5 300

831 3 300

997.2 2.5 300

1246.5 2 300

1662 1.5 300

2493 1 300

When we held the temperature constant and cut the volume in half, each time we reduced the volume we observed that the pressure (increased/decreased) by a factor of (1 / 2 / 4 /8 / 10). Circle the correct answer.

When we held the temperature constant and reduced the volume from 10 to 1, we observed that the pressure (increased/decreased) by a factor of ( 1 / 2 / 4 / 8 / 10). Circle the correct answer.

Part II

P (Pa) V (cm^3) T (K)

831 1 100

831 2 200

831 4 400

831 8 800

831 10 1000

When we held the pressure constant and double the temperature, each time we doubled the temperature we observed that the volume (increased/decreased) by a factor of ( 1 / 2 / 4 / 8 / 10). Circle the correct answer.

When we held the pressure constant and increased the temperature from 100 to 1000, we observed that the volume (increased/decreased) by a factor of (1 / 2 / 4 / 8 / 10). Circle the correct answer.

Part III

P (Pa) V (cm^3) T (K)

166.2 5 100

332.4 5 200

664.8 5 400

1329.6 5 800

1662 5 1000

When we held the volume constant and doubled the temperature, each time we doubled the temperature we observed that the pressure (increased/decreased) by a factor of (1 / 2 / 4 / 8 / 10). Circle the correct answer.

When we held the volume constant and increased the temperature from 100 to 1000, we observed that the pressure (increased/decreased) by a factor of (1 / 2 / 4 / 8 / 10). Circle the correct answer.

Part IV

(a.) We observed that as we compress the gas (decrease the volume) the temperature (increased/decreased) and pressure (increased/decreased). Circle the correct answers.

(b.) We observed that as we compress the gas (decrease the volume) the molecules moved (faster/slower) Circle the correct answer.

(c.) As we compressed the gas the temperature (increased/decreased) and kinetic energy of the molecules (increased/decreased) and the molecular activity (increased/decreased). Circle the correct answers.

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