A $5 L$ $5 L$ container holds $12$ $12$ mol and $9$ $9$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $420^{o} K$ $420 ^oK$ . By how much does the pressure change?

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A $5 L$ $5 L$ container holds $12$ $12$ mol and $9$ $9$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $420^{o} K$ $420 ^oK$ . By how much does the pressure change?

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Answer 1 · 2018-07-06T06:14:11

the pressure change from 126 Bar to 12 Bar

Explanation:

the reaction is
$3 B + 4 A = B_{3} A_{4} + Q$ $3 B + 4 A =B_3A_4 + Q$
so if at the beginning you have (12+9) = 21 mol of gases, after the reaction you have 3 mol of gas( we don't know if we have still gas?)
at the beginning with the gases'law, we have:
$P = n R \frac{T}{V} = 21 m o l \times 8, 31 \frac{J}{m o l \times K} \times \frac{360 K}{0, 005 m^{3}} = 126 \times 10^{5} \frac{J}{m^{3}} = 126 \times 10^{5} \frac{N}{m^{2}} = 126 \times 10^{5} P a s c a l = 126 B a r$ $P= nRT/V= 21 mol xx 8,31 J/(mol xx K) xx (360K)/(0,005m^3)=126 xx 10^5 J/m^3 = 126 xx 10^5 N/m^2 = 126 xx 10^5 Pascal = 126 Bar$
after the reaction:
$P = n R \frac{T}{V} = 3 m o l \times 8, 31 \frac{J}{m o l \times K} \times \frac{420 K}{0, 005 m^{3}} = 12 \times 10^{5} \frac{J}{m^{3}} = 12 \times 10^{5} \frac{N}{m^{2}} = 12 \times 10^{5} P a s c a l = 12 B a r$ $P= nRT/V= 3 mol xx 8,31 J/(mol xx K) xx (420K)/(0,005m^3)=12 xx 10^5 J/m^3 = 12 xx 10^5 N/m^2 = 12 xx 10^5 Pascal = 12 Bar$

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A $5 L$ $5 L$ container holds $12$ $12$ mol and $9$ $9$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $420^{o} K$ $420 ^oK$ . By how much does the pressure change?

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Explanation:

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A 5L5 L container holds 1212 mol and 99 mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from 360oK360^oK to 420oK420 ^oK. By how much does the pressure change?

1 Answer

Explanation:

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A $5 L$ $5 L$ container holds $12$ $12$ mol and $9$ $9$ mol of gasses A and B, respectively. Every three of molecules of gas B bind to four molecule of gas A and the reaction changes the temperature from $360^{o} K$ $360^oK$ to $420^{o} K$ $420 ^oK$ . By how much does the pressure change?