Respuesta :
Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83 [tex]\frac{KJ}{kg}[/tex]
Explanation:
Given Data
Initial pressure [tex]P_{1}[/tex] = 100 k pa
Initial temperature [tex]T_{1}[/tex] = 25 degree Celsius = 298 Kelvin
Final pressure [tex]P_{2}[/tex] = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ [tex]V_{1}[/tex] = [tex]V_{2}[/tex] ------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ [tex]\frac{P_{2} }{P_{1}}[/tex] = [tex]\frac{T_{2} }{T_{1}}[/tex]
⇒ Put all the values in the above formula we get the final temperature
⇒ [tex]T_{2}[/tex] = [tex]\frac{300}{100}[/tex] × 298
⇒ [tex]T_{2}[/tex] = 894 Kelvin
(A). Work done during the process is given by W = P × ([tex]V_{2} -V _{1}[/tex])
From equation (1), [tex]V_{1}[/tex] = [tex]V_{2}[/tex] so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m [tex]C_{v}[/tex] ( [tex]T_{2}[/tex] -[tex]T_{1}[/tex] )
Where m = mass of the gas = 1 kg
[tex]C_{v}[/tex] = specific heat at constant volume of nitrogen = 0.743 [tex]\frac{KJ}{kg k}[/tex]
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83 [tex]\frac{KJ}{kg}[/tex]
Therefore the value of heat transfer during the process Q = 442.83 [tex]\frac{KJ}{kg}[/tex]
