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Volgvan
4 years ago
6

What causes the force that produces gas pressure?

Chemistry
1 answer:
Yuliya22 [10]4 years ago
3 0

( The rapid motion and collisions of molecules ) with the walls of a container causes pressure.

Source; https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Gases/Gas_Pressure

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For a given compound, list the decreasing order of entropy for a liquid, solid, and gas.
Svetradugi [14.3K]

Answer:

B. gas>liquid>solid

Explanation:

Entropy increases with increase in randomness,in gas randomness is higher than in liquid than in solid!

✌️:)

3 0
3 years ago
Why aren't hydrogen fuel cells currently available to the public​
Romashka-Z-Leto [24]

Answer:

they require more testing in real-word conditions

Explanation:

One of the biggest reasons for this very slow uptake of hydrogen vehicles is the existing infrastructure. ... The sceptics' first argument against hydrogen vehicles is that they're less efficient than EVs are. Because hydrogen doesn't occur naturally, it has to be extracted, then compressed in fuel tanks.

8 0
3 years ago
Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar
kakasveta [241]

Answer:

The answer is "\bold{0.525\ \ atm^{-1}}"

Explanation:

Given equation:

2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)

Given value:

\Delta rH =-198.2 \ \ \frac{KJ}{mol}

Kp=1100 \ K

\Delta x = 2-(2+1)\\\\

     = 2-(2+1)\\\\= 2-(3)\\\\= -1

\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm)     &1-2x&0.5-x&2x\end{array}\right

calculating the total pressure on equilibrium=  (1-2x)+(0.5-x)+2x \ atm\\\\

                                                                         = 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\

\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to  x= 0.15\\\\

calculating the pressure in  So_2:

= (1-2 \times 0.15)

= 1-0.30 \\\\ =0.70 \ atm

calculating the pressure in  O_2:

= (0.5- 0.15)\\\\= 0.35 \ atm \\

calculating the pressure in  So_3:

= (2 \times 0.15)\\\\= (.30) \ atm \\\\

Calculating the Kp at 1100 K:

= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\=  0.5247 \ \  or \ \  0.525 \ \ atm^{-1}  \\\\

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