Heat added to Monatomic Gas at Constant Pressure

In summary, to find the change in temperature of a monatomic gas at constant pressure, you simply solve for delta T using the equation 650J = (5/2)(21)(8.31 J/mol.K)(delta T). The result should be rounded to 3 digits and will be in Kelvins, as the molar heat capacity is temperature independent. It is also important to note that for a monatomic gas, U = 3nRT/2, not nRT.
  • #1
KendrickLamar
27
0

Homework Statement



If 650 J of heat are added to 21 moles of a monatomic gas at constant pressure, how much does the temperature of the gas increase? (in Kelvins)

Homework Equations


U = nRT
Q=(5/2)nR(T2-T1)


The Attempt at a Solution



well how do you even know how much the temperature changes by if they don't give u an initial temperature? or is it just as simple as solving for delta T in which you just do

650J = (5/2)(21)( 8.31 J/mol.K )(Delta T)

is it really that simple or do i have to do something with the 21 moles and convert it or anything? and once i do get the Delta T must that be converted or something in order to have it end up in Kelvins? I ended up with 1.4899K I am not sure if that's right though.
 
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  • #2
It is correct. Do not write it out with more than 3 digits. ehild
 
  • #3
KendrickLamar said:

Homework Statement



If 650 J of heat are added to 21 moles of a monatomic gas at constant pressure, how much does the temperature of the gas increase? (in Kelvins)

Homework Equations


U = nRT
Careful. For a monatomic ideal gas, internal energy, U = 3nRT/2
Q=(5/2)nR(T2-T1)
Note that T is in Kelvins, and Q is in Joules and R is in Joule/mol Kelvin. The molar heat capacity Cp (=5R/2) is temperature independent ie. it is the same for all T.

The Attempt at a Solution



well how do you even know how much the temperature changes by if they don't give u an initial temperature? or is it just as simple as solving for delta T in which you just do

650J = (5/2)(21)( 8.31 J/mol.K )(Delta T)

is it really that simple ...
It is that simple.

AM
 
  • #4
wait so is it 3/2 or 5/2? because i have 2 equations from the formula sheet he gave us one says

Cp = 5/2 * R and one says Cv = 3/2 * R ? or in this case is it the 3/2*5/2 or something?
 
  • #5
Cp is the molar specific heat capacity at constant pressure, and it is 5/2 R for mono-atomic gases. You add heat at constant pressure, so Cp was correctly used.

ehild
 
  • #6
thank you guys, appreciate it alot
 
  • #7
KendrickLamar said:
wait so is it 3/2 or 5/2? because i have 2 equations from the formula sheet he gave us one says

Cp = 5/2 * R and one says Cv = 3/2 * R ? or in this case is it the 3/2*5/2 or something?
Cp = 5R/2 and Cv=3R/2. But you had U = nRT. That is not correct. U = nCvT = (3/2)nRT.

AM
 

1. What is the significance of adding heat to a monatomic gas at constant pressure?

The addition of heat to a monatomic gas at constant pressure allows for an increase in the internal energy of the gas, resulting in an increase in its temperature.

2. How does the heat added affect the volume of the monatomic gas?

The volume of the monatomic gas will increase as the gas molecules absorb the added heat energy and move more rapidly, thus expanding the gas and occupying more space.

3. Is the temperature change proportional to the amount of heat added?

Yes, the temperature change in a monatomic gas at constant pressure is directly proportional to the amount of heat added. This relationship is known as the specific heat capacity of the gas.

4. How does the pressure of the gas change when heat is added at constant pressure?

The pressure of the gas will also increase as the gas molecules gain kinetic energy and collide more frequently with the walls of the container, exerting a greater force per unit area.

5. What is the equation for calculating the change in internal energy of a monatomic gas at constant pressure?

The change in internal energy, denoted as ΔU, can be calculated using the equation ΔU = nCVΔT, where n is the number of moles of gas, CV is the specific heat capacity at constant volume, and ΔT is the change in temperature.

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