Chemistry, 03.09.2020 02:01 justinjoyner12p5ox1r

In a chemical processing plant, a reaction chamber of fixed volume V0 is connected to a reservoir chamber of fixed volume 4V0 by a passage containing a thermally insulating porous plug. The plug permits the chambers to be at different temperatures. The plug allows gas to pass from either chamber to the other, ensuring that the pressure is the same in both. At one point in the processing, both chambers contain gas at a pressure of 1.00 atm and a temperature of 27.0°C. Intake and exhaust valves to the pair of chambers are closed. The reservoir is maintained at 27.0°C while the reaction chamber is heated to 400°C. What is the pressure in both chambers after that is done? 50. Why is the following situation impossible?

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Answer from: csciahetano

Explanation:

The two chambers are at temperature of 27° and pressure of 1 atm . Now temperature of second chamber is increased to 400°C so its pressure will be increased but pressure of chamber whose temperature remains at 27°C will remain unaltered . Hence the pressure of chamber at 27°C will be at 1 atm.

The volume of second chamber is 4 V₀ , temperature changed from 27 + 273 = 300 k to 400 + 273 = 673 K , initial pressure was 1 atm . Let final pressure be P

From gas law formula

1 x 4 V₀ / 300 = P x 4 V₀ / 673

P = 1 x 673 / 300

= 2.24 atm .

The pressure of the chamber at 400°C will be 2.24 atm.

Answer from: Quest

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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