Answer:
The temperature will be greater than 25°C
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C
10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.
Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.