Respuesta :
Explanation:
Given that,
Pressure = 3.94 Mpa
Benzene data
[tex]T_{c}=564 K[/tex]
[tex]P_{cr}=4.92\ MPa[/tex]
[tex]R=0.1064\ kJ/kg.K[/tex]
[tex]C_{p}=0.502+0.00252\ T KJ/kg.K[/tex]
(a). We need to calculate the temperature
Using equation of ideal gas at constant volume
[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]T_{1}=\dfrac{P_{1}T_{2}}{P_{2}}[/tex]
Put the value into the formula
[tex]T_{1}=\dfrac{564\times3.94 }{4.92}[/tex]
[tex]T_{1}=451.6\ K[/tex]
(b). We need to calculate the work done
Here, the work done by the system is zero because this process is isochoric process in which the volume is constant.
(c). We need to calculate the heat transferred to the benzene during this process
Using formula of heat transferred
[tex]\Delta Q=C_{v}dT[/tex]
[tex]\Delta Q=(C_{p}-R)dT[/tex]
[tex]\Delta Q=(C_{p}-R)\times(T_{2}-T_{1})[/tex]
[tex]\Delta Q=(0.502-0.1064)\times((345+273)-451.6)[/tex]
[tex]\Delta Q=65.82\ KJ/kg[/tex]
Hence, This is the required solution.
