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Ainat [17]
2 years ago
10

Prepare a stock available at home following the procedure you learned in this lesson. Have a picture of the finished product and

a narration on how you do it​

Physics
1 answer:
Bingel [31]2 years ago
6 0

Answer:

NO, 1. is stocks

2. is also stocks

bouquit GARNI IS NUMBER 3

NUMBER 4 IS ACID PRODUCTS

number 5 is brown stock

Explanation:

stocks are bones

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A mole of ideal gas expands at T=27 °C. The pressure changes from 20 atm to 1 atm. What’s the work that the gas has done and wha
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Answer:

  • The work made by the gas is 7475.69 joules
  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

E= c_v n R T

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By the first law of thermodynamics, we know

\Delta E = \Delta Q - \Delta W

where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

\Delta Q = \Delta W

7 0
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