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DochEvi [55]
3 years ago
12

When the temperature of a closed vessel containing a gas is increased by 273 °C, the pressure was found to increase by 20%. What

is the initial temperature of the closed vessel?
\\  \\
No spamming or Irrelevant Answer.
Ty!​
Chemistry
2 answers:
Anon25 [30]3 years ago
8 0

Answer:

<h3>Hey There!</h3>

<em>The answer is 1365k </em>

Explanation:

<em>Refer to the attached picture.</em>

\\  \\  \\

Hope it Helps:)

~ƒöᏝϗʆѻʁᶥąռ¹₃

Rus_ich [418]3 years ago
5 0

Answer:

Solution given;

1st temperature [t1]=T

initial temperature [t2]=T+273°C

1st pressure [P]=P

and

initial pressure[P2]=P+20%of P=120/100V=1.2P

now

we have

According to Gay lussac's law

P\alpha T

\frac{P}{T}=K where k is constant

at 2 different states

\frac{P1}{T1}=\frac{P2}{T2}

  • \frac{P}{T}=\frac{1.2P}{T+273}

\frac{T+273}{T}=\frac{1.2P}{P}

doing crisscrossed multiplication

T+273=1.2T

273=1.2T-T

0.2T=273

T=\frac{273}{0.2}=1365°C or 1365K

<u>I</u><u>nitial temperature of the closed </u><u>vessel</u><u> </u><u>i</u><u>s</u><u> </u><u>1365°C or 1365</u><u>K</u><u>.</u>

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