Respuesta :
Answer : The value of [tex]q\text{ and }\Delta U[/tex] is 286.2 J and 286.2 J respectively.
Explanation : Given,
Moles of sample = 0.877 mol
Change in temperature = 15.7 K
First we have to calculate the heat absorbed by the system.
Formula used :
[tex]q=n\times c_v\times \Delta T[/tex]
where,
q = heat absorbed by the system = ?
n = moles of sample = 0.877 mol
[tex]\Delta T[/tex] = Change in temperature = 15.7 K
[tex]c_v[/tex] = heat capacity at constant volume of [tex]N_2[/tex] (diatomic molecule) = [tex]\frac{5}{2}R[/tex]
R = gas constant = 8.314 J/mol.K
Now put all the given value in the above formula, we get:
[tex]q=0.877mol
\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K[/tex]
[tex]q=286.2J[/tex]
Now we have to calculate the change in internal energy of the system.
[tex]\Delta U=q+w[/tex]
As we know that, work done is zero at constant volume. So,
[tex]\Delta U=q=286.2J[/tex]
Therefore, the value of [tex]q\text{ and }\Delta U[/tex] is 286.2 J and 286.2 J respectively.
The value of q is 28.2 J while the value of ΔT is 286.2 J
The heat capacity of an object will reveal to us the exact amount of heat needed to raise a specific amount of it by one degree.
Also, the value of the capacity of an object is based on how the heat is added which can be by either constant pressure or constant volume.
Further Explanation
From the given question, the heat capacity is at constant volume, therefore the formula to be use is Q = nCνΔT.
That is:
Q = n x Cν x ΔT.
- Where Q is the heat absorbed at constant volume
- N is the moles of the sample
- ΔT is the change in temperature
- Cν is the heat capacity at constant volume (recall N₂ is a diatomic molecules)
The value of Q =?
n = 0.877 mol
ΔT = 15.7 K
Cν = 5/2 R (N₂ diatomic molecules), where the value for R is 8.314 J.
If you substitute the value, then we get:
Q = 0.877 mol x 5/2 x 8.314J/mol.K x 15.7K
Q = 286.2 J
To calculate the change in the thermal energy, you should us the formula
ΔU = q + w
ΔU = q + (0) (work done at constant volume is zero)
ΔU = 286.2 J + 0
ΔU = 286.2 J
Therefore, the value of q is 28.2 J while the value of ΔT is 286.2 J
LEARN MORE:
- A 0.773 mol sample of Xe(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the temperature of the gas by 20.1 K https://brainly.com/question/13829259
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KEYWORDS:
- temperature change
- heat capacity
- internal heat
- N₂
- thermal energy
- constant volume
- constant pressure
