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KATRIN_1 [288]
2 years ago
14

In a physics lab, you measure the vibrational-rotational spectrum of hcl. the estimated separation between absorption peaks is:_

________-
Physics
1 answer:
BlackZzzverrR [31]2 years ago
8 0

The appropriate value in blank given is Δf = 5.5 x 10^{11} Hertz.

We have vibrational - rotational spectrum Hydrochloric Acid.

We have to investigate the estimated separation between absorption peaks and fill the blank.

<h3>What is vibrational - rotational spectrum ?</h3>

Rotational–vibrational spectroscopy is a branch of molecular spectroscopy. It deals with the infrared and Raman spectra of molecules in the gaseous phase.

According to the question -

The estimated separation between absorption peaks in the vibrational-rotational spectrum of HCl is denoted by Δf and is equal to -

Δf = 5.5 x 10^{11} Hertz

Hence, the appropriate value in blank given is Δf = 5.5 x 10^{11} Hertz.

To learn more about vibrational-rotational spectrum, visit the link below-

brainly.com/question/18403840

#SPJ4

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Air enters a turbine operating at steady state at 8 bar, 1400 K and expands to 0.8 bar. The turbine is well insulated, and kinet
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To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

Where

T_1 =Temperature at inlet of turbine

T_2 = Temperature at exit of turbine

P_1 = Pressure at exit of turbine

P_2 =Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

m_i = m_0 = m

m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W

Where,

m = mass

m_i = mass at inlet

m_0= Mass at outlet

h_i = Enthalpy at inlet

h_0 = Enthalpy at outlet

W = Work done

Q = Heat transferred

V_i = Velocity at inlet

V_0= Velocity at outlet

Z_i= Height at inlet

Z_0= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

h_i = h_0 + W

W = h_i -h_0

Using the relation T-P we can find the final temperature:

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}

\frac{T_2}{1400K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}

T_2 = 725.126K

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

So:

W = h_i -h_0

W = C_p (T_1-T_2)

W = 1.005(1400-725.126)

W = 678.248kJ/Kg

Therefore the maximum theoretical work that could be developed by the turbine is 678.248kJ/kg

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4 years ago
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