combined gas law assignment

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Combined Gas Law – Definition, Formula, Examples

The combined gas law is an ideal gas law that combines Boyle’s law , Charles’s law , and Gay-Lussac’s law . It states the the ratio between the pressure-volume product and absolute temperature of a gas is a constant. Pressure, volume, and temperature are allowed to change, but the amount of gas (number of moles) remains unchanged. Basically, the combined gas law is the ideal gas law, except that it’s missing Avogadro’s law . The combined gas law doesn’t have an official discoverer, so it doesn’t have a name.

• Combined gas law: PV/T = k
• Boyle’s law: PV = k
• Charles’s law: V/T = k
• Gay-Lussac’s law: P/T = k

Combined Gas Law Formula

The basic formula for the combined gas law related pressure (P), volume (V), and absolute temperature (T):

The constant k is a true constant, so long as the number of moles of a gas remains the same. If the amount of gas varies, then k changes.

The practical formula for the combined gas law gives “before and after” conditions of a gas:

P 1 V 1  / T 1  = P 2 V 2  / T 2

Rearranging the variables gives:

P 1 V 1  T 2  = P 2 V 2  T 1

Any units of pressure and volume are fine, but temperature must be absolute. In other words, convert Fahrenheit and Celsius temperatures to Kelvin.

How the Combined Gas Law Applies to Everyday Life

The combined gas law has practical applications in everyday life. It applies whenever the amount of gas remains constant, but pressure, volume, and temperature change. For example, the law predicts the behavior of cloud formation, refrigerators, and air conditioners. It’s also used in other thermodynamics and fluid mechanics calculations.

Because the combined gas law is an ideal gas law, it only approximates the behavior of real gases . The law becomes less accurate at high pressures and temperatures.

Combined Gas Law Example Problem

Find the volume of a gas at 760.0 mm Hg pressure and 273 K when 2.00 liters is collected at 745.0 mm Hg and 25.0 °C.

First convert 25.0 °C to the Kelvin scale . This gives you 298 Kelvin.

Next, plug the values into the combined gas law formula. The most common mistake students make is mixing up which numbers go together. Writing down what you’re given helps avoid this error:

P 1  = 745.0 mm Hg V 1  = 2.00 L T 1  = 298 K P 2  = 760.0 mm Hg V 2  = x (the unknown you’re solving for) T 2  = 273 K

Arrange the formula to solve for the unknown:

P 1 V 1  / T 1  = P 2 V 2  / T 2 P 1 V 1 T 2  = P 2 V 2 T 1 V 2  = (P 1 V 1 T 2 ) / (P 2 T 1 )

Plug in the numbers:

V 2  = (745.0 mm Hg · 2.00 L · 273 K) / (760 mm Hg · 298 K) V 2  = 1.796 L V 2  = 1.80 L

• Castka, Joseph F.; Metcalfe, H. Clark; Davis, Raymond E.; Williams, John E. (2002).  Modern Chemistry . Holt, Rinehart and Winston. ISBN 978-0-03-056537-3.
• Fullick, P. (1994).  Physics . Heinemann. ISBN 978-0-435-57078-1.
• Raff, Lionel M. (2001) Principles of Physical Chemistry (1st ed.). Pearson College Div. ISBN:  978-0130278050.

Related Posts

(P 1 V 1 ) / T 1 = (P 2 V 2 ) / T 2

V 2 = (P 1 V 1 T 2 ) / (T 1 P 2 )

P 1 = 300.0 torr P 2 = 600.0 torr V 1 = 800.0 mL      V 2 = x T 1 = 250. K T 2 = 500. K

    P 1 V 1 T 2 V 2 = ––––––     P 2 T 1

    (300.0 torr) (800.0 mL) (500.0 K) V 2 = –––––––––––––––––––––––––––     (250.0 K) (600.0 torr) V 2 = 800.0 mL

V 2 = [(0.92105) (500) (293)] / [(473) (30)] V 2 = 9.51 L

P total = P O 2 + P H 2 O P O 2 = P total - P H 2 O P O 2 = 725.0 mmHg - 25.2 mmHg = 699.8 mmHg The 25.2 value came from here . I looked up the value associated with 26.0 °C and converted it from kPa to mmHg following the instructions given.

P 1 = 699.8 mmHg P 2 = 800.0 mmHg V 1 = 690.0 mL      V 2 = x T 1 = 299.0 K T 2 = 325.0 K A common student error is to use Dalton's Law, but then use the total pressure value in the combined gas law instead of using the correct value. The correct pressure to use for P 1 is the 699.8 value, not the 725 value. The 725 is the pressure of an oxygen/water mixture and we want ONLY the oxygen (which is the 699.8 value).

x = [(699.8) (690.0) (325)] / [(299) (800.0)] x = 656 mL (to three sig figs)

P 1 = 0.250 atm P 2 = 2.00 atm V 1 = 300.0 L      V 2 = x T 1 = 400.0 K T 2 = 200.0 K Note how the problem statement is worded so as to give the starting values last.

V 2 = (P 1 V 1 T 2 ) / (T 1 P 2 ) x = [(0.25) (300) (200)] / [(400) (2)] x = 18.75 L To three sig figs, this is 18.8 L

V 2 = [(785 mmHg) (45.5 mL) (303 K)] / [(288 K) (745 mmHg)] V 2 = 50.3757 mL To three sig figs, the answer is 50.4 mL

P 1 = 760.0 mmHg P 2 = 360.0 mmHg V 1 = 400.0 L      V 2 = x T 1 = 295 K T 2 = 303 K V 2 = [(760 mmHg) (400 mL) (303 K)] / [(295 K) (360 mmHg)] V 2 = 867 mL (to three sig figs)

(a) What is the partial pressure of H 2 ? (b) What is the partial pressure of H 2 O? (c) What is the volume of DRY hydrogen at STP?

P total = P H 2 + P H 2 O P H 2 = P total - P H 2 O 740.0 - 15.5 = 724.5 mmHg I used a different reference source than previously used for the vapor pressure of water. There are many available on the Internet.

    P 1 V 1 T 2 V 2 = ––––––     P 2 T 1     (724.5 mmHg) (400.0 mL) (273 K) V 2 = ––––––––––––––––––––––––––––     (291 K) (760.0 mmHg) V 2 = 358 mL (to three sig figs)

Let us assign P 1 = 1, therefore P 2 = 0.75 Let us assign V 1 = 1, therefore V 2 = 1.4 I won't bother with units on P or V. Your teacher may want the units added in, so I'll do that below. T 1 = −10 °C = 263 K P 1 V 1 /T 1 = P 2 V 2 /T 2 [(1 atm) (1 L)] / 263 K = [(0.75 atm) (1.4 L)] / x (1 atm) (1 L) (x) = (263 K) (0.75 atm) (1.4 L) x = 276.15 K = 3.15 °C

P 1 = 3.00 atm P 2 = 1.00 atm V 1 = 8.06 L V 2 = x T 1 = 2.00 K T 2 = 1.00 K Note the made up values for P and T.

P 1 V 1 / T 1 = P 2 V 2 / T 2 [(3.00 atm) (8.06 L)] / 2.00 K = [(1.00 atm) (x)] / 1.00 K x = 12.1 L (to three sig figs)

P 1 = 0.939 atm P 2 = 1.00 atm V 1 = 9.40 L V 2 = 10.0 L T 1 = x T 2 = 298 K Note how the problem gives you the ending conditions regarding PVT and asks you for a starting condition. Also, note that temperature is asked for. Compare this to the usual case of asking for the final volume.

P 1 V 1   P 2 V 2 ––––– = ––––– T 1   T 2 T 1 P 2 V 2 = P 1 V 1 T 2     P 1 V 1 T 2 T 1 = ––––––     P 2 V 2

    (0.939 atm) (9.40 L) (298 K) T 1 = ––––––––––––––––––––––––     (1.00 atm) (10.0 L) T 1 = 263 K The form of the final temperature was not specificed, but it is usually asked for in Celsius, so: T 1 = −10. °C

P 1 V 1   P 2 V 2 ––––– = ––––– n 1 T 1   n 2 T 2

P 1 = 1.00 atm P 2 = 2.00 atm V 1 = 8.00 mL V 2 = x n 1 = 1.00 mol n 2 = 0.85 mol T 1 = 1.00 K T 2 = 4.00 K By the way, having a gas at 1.00 K is pretty much an impossible thing. Pure helium-3 liquifies at about 3.2 K. No gas exists as a gas at 1 K. The point, of course, is to make the absolute temperature quadruple. We could use 100 K and 400 K and get the same answer. Or use 200 K and 800 K. The key point is that the temperature quadruples. And, note that it is the absolute temperature (in K) that quadruples, not the temperature in degrees Celsius. Notice how I interpreted the 15% to mean 15% of the moles of gas escaped. If I decided that 15% of the mass escaped, the problem answer would be the same. I will leave it to you to figure out, if you so desire.

(1.00 atm) (8.00 mL)   (2.00 atm) (x) ––––––––––––––––– = ––––––––––––––– (1.00 mol) (1.00 K)   (0.85 mol) (4.00 K)

(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) = (2.00 atm) (x) (1.00 mol) (1.00 K)

    (1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) x = –––––––––––––––––––––––––––––––––     (2.00 atm) (1.00 mol) (1.00 K) x = 13.6 mL

P 1 = 733.4 mmHg      P 2 = 760.0 mmHg      V 1 = 1.00 L      V 2 = x      T 1 = 298 K      T 2 = 273 K     

(733.4 mmHg) (1.00 L)   (760.0 mmHg) (x) –––––––––––––––––––   =   ––––––––––––––– 298 K   273 K After a bit of math, we find: x = 0.884 L

0.672 g / 0.884 L = 0.760 g/L

(a) The pressure is tripled while the absolute temperature is doubled. (b) The absolute temperature is doubled while the pressure is reduced by half. (c) The pressure and the absolute temperature are both doubled. (d) The temperature is increased by four times while at the same time the pressure is doubled.

P 1 = 1 kPa P 2 = 3 kPa V 1 = 1 L V 2 = ??? T 1 = 1 K T 2 = 2 K Note how the pressure tripled (from 1 to 3) and the temperature doubled (from 1 to 2).

V 2 = (P 1 V 1 T 2 ) / (P 2 T 1 ) Notice I isolated V 2 V 2 = [(1) (1) (2)] / [(3) (1)] V 2 = 0.67 L In other words, V 2 is two-thirds of V 1 .

V 2 = (P 1 V 1 T 2 ) / (P 2 T 1 )

V 2 = (P 1 V 1 (2) / (P 2 (1)

V 2 = (2) V 1 (2) / (1) (1) Notice how I went from 2 to 1. I felt that was clearer than going from 1 to 0.5.

V 2 = 4V 1 The volume is increased by a factor of 4.

V 2 = (P 1 V 1 T 2 ) / (P 2 T 1 ) V 2 = (1) V 2 (2) / (2) (1) Note placement of values for temperature, with T 1 in the denominator and T 2 in the numerator. With the symbolic equation being all on one line, you might be tempted to think the temperature was cut in half. Not so! V 2 = V 1

P 1 = 1 kPa P 2 = 2 kPa V 1 = 1 L V 2 = ??? T 1 = 1 K T 2 = 4 K

V 2 = (P 1 V 1 T 2 ) / (P 2 T 1 ) V 2 = (1) V 1 (4) / (2) (1) V 2 = 2V 1 In other words, the volume doubles.

25.0 °C and 100.0 kPa You may find more discussion here.

P 1 = 101.325 kPa P 2 = 100.0 kPa V 1 = 22.414 L V 2 = ??? T 1 = 273 K T 2 = 298 K Note the use of values for STP and molar volume at STP.

P 1 V 1   P 2 V 2 ––––– = ––––– T 1   T 2 (101.325 kPa) (22.414 L)   (100.0 kPa) (V 2 ) ––––––––––––––––––––– = –––––––––––––– 273 K   298 K V 2 = [(101.325 kPa) (22.414 L) (298 K)] / [(100.0 kPa) (273 K)] V 2 = 24.8 L (to three sig figs)

PV = nRT (100.0 kPa / 101.325 kPa/atm) (V) = (1.00 mol) (0.08206 L atm / mol K) (298 K) V = 24.8 L

Here's the cross-multiplied form of the combined gas law: P 1 V 1 T 2 = P 2 V 2 T 1 (102.5 kPa) (0.730 dm 3 ) (273 K) = (101.3 kPa) (V 2 ) (294 K) V 2 = 0.685887 dm 3

0.685887 dm 3 times 0.900 g/dm 3 = 0.617 g

(a) Is it possible to change the temperature of the gas at the same time such that the volume of the gas doesn't change? (b) If yes, calculate the new temperature of the gas

We can't answer this without doing the calculation for (b). This is because the temperature must go down to keep the volume at 10.0 m 3 . If the temperature most go down to absolute zero (or below), then the answer would to (a) would be no. Otherwise, we would answer yes. I know the temp must go down because of this: the pressure went down, therefore the volume went up (assuming constant temperature). In order to get the volume back to 10, we must cool the gas down (assuming constant pressure).

P 1 V 1   P 2 V 2 ––––– = ––––– T 1   T 2 (2 atm) (10.0 m 3 )   (1 atm) (10.0 m 3 ) ––––––––––––– = ––––––––––––– 285 K   x I used arbitrary values for the pressure. The actual values don't matter, just as long as the pressure gets cut in half.

2 atm   1 atm ––––– = ––––– 285 K   x x = 142.5 K

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Chemistry: Gas Laws

Influence of Parameter Deviations Vis-à-Vis Assignment Schedule on Thermally Stressed State of Main Thermal Power Plant Equipment

• Published: 23 May 2024
• Volume 57 , pages 918–921, ( 2024 )

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• Yu.A. Radin 1 &
• T. S. Kontorovich 1  

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The current guidelines recommend evaluating the quality of start-up and shut-down modes, as well as operation modes under load, according to the compliance of start-up parameters with assignment schedules, standard instructions, performance charts, and reliability criteria and their limit values. The quality of maintaining operating modes is assessed via the permissible deviations of parameters from the values specified in standard instructions or assignment schedules. This study analyzes the approach and the permissible parameter deviations, according to which the start-up operations quality is assessed in terms of thermally stressed state of the critical elements of a thermal scheme. The analysis was performed on the basis of the thermally stressed state calculations of the T-125/150-7.7 turbine’s high-pressure rotor and the PGU-450 waste heat boiler’s high-pressure superheater outlet header with deviations of live steam temperature from the assignment schedule. Calculations were performed using the software package ANSYS. It is proposed to evaluate the quality of start-up modes on the basis of calculation of the cyclic strength of the power plant critical elements and not via deviation of the control parameter values without considering the specific time of their occurrence in relation to the assignment schedule, temperature state of element under consideration, and steam flow, among others.

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The effectiveness of combined-cycle power plants hot startups simulating

Condition Monitoring of Heat Recovery Steam Boilers as a Basis for Elaborating Process Regulations for Their Operation

Methodical fundamentals for constructing startup assignment schedules for combined-cycle power plants considering damage accumulation, explore related subjects.

• Geoengineering

RD 153-34.0-20.585-00. Guidelines for Analyzing the Quality of Start-Up (Shutdown) of the Main Thermal Power Plant Equipment [in Russian], approved by the RJSC “UES of Russia” 12/28/99, SPO ORGRES, Moscow (2000).

RTM 108.021.1093–85. Parts of Stationary Steam Turbines. Calculation for Low-Cycle Fatigue [in Russian], approved on 09/13/1985, in force since 07/01/1986, NPO CKTI, Leningrad (1985).

E. R. Plotkin and A. Sh. Leizerovich, Starting Modes of Steam Turbines of Power Units [in Russian], Énergiya, Moscow (1980).

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L. Ya. Liberman and M. N. Peisikhis, Properties of Steels and Alloys Used in Boiler and Turbine Construction. Guidelines [in Russian], CKTI, Leningrad (1966).

A. S. Zubchenko, Grade Guide of Steels and Alloys [in Russian], Mashinostroenie, Moscow (2003).

B. E. Neimark, Physical Properties of Steels and Alloys Used in Energy. Guidebook [in Russian], Énergiya, Moscow – Leningrad (1967).

STO 70238424.27.100.016–2009. Combined-Cycle Plants. Organization of Operation and Maintenance. Standards and Requirements [in Russian], in force since 01/29/2009, NP “INVEL,” Moscow (2009).

M. A. Mikheev and I. M. Mikheeva, Fundamentals of Heat Transfer [in Russian], Énergiya, Moscow (1977).

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Mosenergo PJSC, Moscow, Russia

Yu.A. Radin & T. S. Kontorovich

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Correspondence to Yu.A. Radin .

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Translated from Élektricheskie Stantsii , No. 10, October 2023, pp. 8 – 12. DOI: https://doi.org/10.34831/EP.2023.1107.10.002

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Radin, Y., Kontorovich, T.S. Influence of Parameter Deviations Vis-à-Vis Assignment Schedule on Thermally Stressed State of Main Thermal Power Plant Equipment. Power Technol Eng 57 , 918–921 (2024). https://doi.org/10.1007/s10749-024-01758-2

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Published : 23 May 2024

Issue Date : March 2024

DOI : https://doi.org/10.1007/s10749-024-01758-2

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