Ideal gas and final temperature

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SUMMARY

The discussion centers on calculating the final temperature of an ideal gas when its internal energy increases by a factor of three. The relevant equations are U = 3/2NkT and U = 3/2nRT, where U represents internal energy, N is the number of molecules, n is the number of moles, k is the Boltzmann constant (1.38 x 10^-23 J/K), and R is the ideal gas constant (8.314 J/mol·K). The initial temperature of the gas is 20 degrees Celsius (293.15 K), and the final temperature can be derived by manipulating these equations and converting the result from Kelvin to Celsius.

PREREQUISITES
  • Understanding of the ideal gas law and internal energy equations
  • Familiarity with temperature conversion between Celsius and Kelvin
  • Knowledge of constants such as Boltzmann's constant (k) and the ideal gas constant (R)
  • Basic algebraic manipulation skills for solving equations
NEXT STEPS
  • Study the derivation and applications of the ideal gas law
  • Learn about the significance of internal energy in thermodynamics
  • Explore temperature conversion methods and their importance in scientific calculations
  • Investigate the implications of varying constants (k and R) based on different gas conditions
USEFUL FOR

This discussion is beneficial for students and professionals in physics and chemistry, particularly those studying thermodynamics and the behavior of gases. It is also useful for anyone involved in calculations related to gas properties and energy transformations.

cathliccat
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I have the question, "Heat is added to an ideal gas at 20 degrees C. If the internal energy of the gas increases by a factor of three - what is the final temperature (in degrees C) round off to the nearest whole number?"

I know U=3/2NkT or U= 3/2nRT
I see I have 3U and I need to convert 20 degrees C to 293.15K, all the other units are the same. I know k = 1.38 X 10^-23 and somewhere it says what R is. I thought if I set the same equation = to each other I could get the answer, 3(3/2 * k * T)= 3/2 * k * T, but I'm not getting anything feasible. What am I doing wrong? Any help is appreciated!

Thanks,
Carissa
 
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In U=3/2nKT, only T is a variable, all other quantities are constant. So if U~T, is you triple the internal energy what happens to T? Well if 3/2nK is a constant, the number of molecules doesn't change so this should be true, what happens to T? That is the question you should be asking yourself.
Cheers,
Ryan
 


First, let's clarify that the equation U = 3/2NkT or U = 3/2nRT is the equation for the internal energy of an ideal gas. This equation assumes that the gas is in a state of constant temperature and volume, and that there are no external forces acting on the gas particles.

Now, to solve the problem, we can use the equation for the internal energy and set it equal to 3 times the original internal energy (since it is stated that the internal energy increases by a factor of three). So we have:

3U = 3(3/2NkT) or 3U = 3(3/2nRT)

Next, we can substitute in the given information that the initial temperature is 20 degrees C (or 293.15K). So the equation becomes:

3U = 3(3/2Nk(293.15)) or 3U = 3(3/2nR(293.15))

Now, we can solve for the final temperature (T) by dividing both sides by 3 and then plugging in the known values for N or n, k, and R:

T = (3U)/(3(3/2)Nk) or T = (3U)/(3(3/2)nR)

Since the question asks for the final temperature in degrees Celsius, we can convert the final temperature from Kelvin to Celsius by subtracting 273.15 from the final temperature in Kelvin.

So the final temperature in degrees Celsius would be:

T = [(3U)/(3(3/2)Nk)] - 273.15 or T = [(3U)/(3(3/2)nR)] - 273.15

Note: The value of R depends on the units used for pressure and volume. In this problem, we are not given information about the pressure or volume, so we cannot determine the exact value of R. However, we can use the value of R for ideal gases, which is 8.314 J/mol·K.

I hope this helps clear up any confusion and leads you to the correct answer. Remember to always double check your units and make sure they are consistent throughout your calculations. Good luck!
 

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