Respuesta :
Answer:
The initial and final temperatures of the gas is 300 K and 600 K.
Explanation:
Given that,
Entropy of the gas = 14.41 J/K
Absorb gas = 6236 J
We know that,
[tex]ds=\dfrac{dQ}{dt}[/tex]
At constant pressure,
[tex]dQ=C_{p}dt[/tex]
[tex]\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}[/tex]
[tex]\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}[/tex]
Put the value into the formula
[tex]14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})[/tex]
[tex]\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}[/tex]
[tex]0.693=ln\dfrac{T_{2}}{T_{1}}[/tex]
[tex]ln2=ln\dfrac{T_{2}}{T_{1}}[/tex]
[tex]T_{2}=2T_{1}[/tex]...(I)
We need to calculate the initial and final temperatures of the gas
Using formula of energy
[tex]\Delta Q=C_{p}\Delta T[/tex]
Put the value into the formula
[tex]6236=2.5\times8.3144(T_{2}-T_{1})[/tex]
[tex]6236=20.786(T_{2}-T_{1})[/tex]
[tex]T_{2}-T_{1}=\dfrac{6236}{20.786}[/tex]
[tex]T_{2}-T_{1}=300[/tex]
Put the value of T₂
[tex]2T_{1}-T_{1}=300[/tex]
[tex]T_{1}=300\ K[/tex]
Put the value of T₁ in equation (I)
[tex]T_{2}=2\times300[/tex]
[tex]T_{2}=600\ K[/tex]
Hence, The initial and final temperatures of the gas is 300 K and 600 K.
