Respuesta :
Answer : The value of [tex]q\text{ and }\Delta U[/tex] is 193.8 J and 193.8 J respectively.
Explanation : Given,
Moles of sample = 0.773 mol
Change in temperature = 20.1 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 = 0.773 mol
[tex]\Delta T[/tex] = Change in temperature = 20.1 K
[tex]c_v[/tex] = heat capacity at constant volume of Xe (mono-atomic molecule) = [tex]\frac{3}{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=0.773mol\times \frac{3}{2}\times 8.314J/mol.K\times 20.1K[/tex]
[tex]q=193.8J[/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=193.8J[/tex]
Therefore, the value of [tex]q\text{ and }\Delta U[/tex] is 193.8 J and 193.8 J respectively.
- The amount of heat (q) requires to raise the temperature at constant volume is 193.76 J.
- The internal energy (U) of the system is 193.76 J.
The amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius is specific heat.
The heat (q) required can be expressed as:
[tex]q=\rm mass\;\times\;specific\;heat\;\times change \;in \;temperature[/tex]
Computation for heat and internal energy
- The moles for the Xe sample have been 0.773 mol.
- The specific heat of Xe sample is 12.471 J/mol.K
- The change in temperature is 20.1 K
The heat required by the sample is given by:
[tex]q=0.773\;\times\;12.471\;\times\;20.1\;\text J\\q=193.76\;\text J[/tex]
The amount of heat requires to raise the temperature at constant volume is 193.76 J.
The internal energy for the reaction at constant volume has work done 0.
The internal energy (U) for the system is given as:
[tex]\rm \textit U=q+work\;done\\\textit U=193.76\;+\;0\;J\\\textit U=193.76\;J[/tex]
The internal energy of the system is 193.76 J.
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