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anygoal [31]
3 years ago
9

Calculate the work (kJ) done during a reaction in which the internal volume contracts from 85 L to 12 L against an outside press

ure of 2.4 atm.
Chemistry
1 answer:
butalik [34]3 years ago
5 0

Answer:

The work done on the system is 17.75_KJ

Explanation:

To solve this question we need to know the required equations from the given variables, thus

Initial volume = 85L

Final volume = 12L

External pressure = 2.4 atm.

work done = - PΔV

2.4×(85-12) = 175.2L×atm

Converting from L•atm to KJ is given by

1 L•atm = 0.1013 kJ

(175.2 L•atm) * (0.1013 kJ / 1 L•atm)

= 17.75 kJ

The work done on the system is 17.75_KJ

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Read 2 more answers
Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

6 0
4 years ago
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