Respuesta :
Answer:
388 K
Explanation:
[tex]T_{i}[/tex] = initial temperature before handle is pushed = 21 C = 21 + 273 = 294 K
[tex]T_{f}[/tex] = final temperature after handle is pushed = ?
[tex]V_{i}[/tex] = initial Volume
[tex]V_{f}[/tex] = final Volume
Given that :
[tex]V_{f} = (0.5) V_{i}[/tex]
For adiabatic process we have
[tex]T_{i} V_{i}^{\gamma -1} = T_{f} V_{f}^{\gamma -1}\\(294) V_{i}^{1.4 -1} = T_{f} ((0.5) V_{i})^{1.4 - 1}\\(294) V_{i}^{1.4 -1} = T_{f} (0.5)^{0.4} V_{i}^{1.4 -1}\\(294) = T_{f} (0.5)^{0.4} \\T_{f} = 388 K[/tex]
Answer:
222.8 K
Explanation:
length of cylinder, l = 20 cm = 0.2 m
diameter of cylinder = 3 cm
radius of cylinder, r = 1.5 cm = 0.015 m
Volume of air initially, V1 = πr²l
V1 = 3.14 x 0.015 x 0.015 x 0.2 = 1.413 x 10^-4 m^3
V2 = V1/2
P1 = 1 atm = 1.01 x 10^5 Pa
T1 = 21°C = 21 + 273 = 294 K
γ = 1.4
This is the adiabatic compression
[tex]T_{1}V_{1}^{1-\gamma }=T_{2}V_{2}^{1-\gamma }[/tex]
[tex]\frac{T_{2}}{T_{1}}=\left ( \frac{V_{1}}{V_{2}} \right )^{1-\gamma }[/tex]
[tex]\frac{T_{2}}{294}=2^{1-1.4 }[/tex]
T2 = 222.8 K
Thus, the temperature of air is 222.8 K.
