Science, asked by mohdizhan2020, 9 months ago

A round bottom flask contains 360 ml of the gas at 29⁰C and 70 cm of Hg. Find the final pressure when it is transferred in a conical flask of 85 ml capacity at a temperature of 57⁰C.​

Answers

Answered by Rajakhavin
3

Answer:

For flask 1:

P1V1=nRT1

or, P1V1T1=nR ...(1)

Similarly, for flask 2

P2V2T2=nR ...(2)

Equating both equations we get,

P1V1T1=P2V2T2P1= 70 cm of Hg =7076=0.9210 atm , P2 = ?V1 = 300 ml , V2 = 75 mlT1=27+273 = 300 K , T2 = 57 + 273 = 330 KNow, 0.9210×300300=P2×75330or, P2 = 4.0526 atm or 307.9 mm of Hg

Answered by sameerronaldo12763
1

Answer:

Final pressure of gas is 205.33 cm of Hg

\mathfrak{\underline{\underline{Explanation:}}}

Initial Conditions

\sf{P_1 = 70 cm \: of \: Hg}\\ \sf{V_1=200ml}\\ \sf{V_1=200ml}\\ \sf{T_1=27°C = 300K}

Final Conditions

\sf{P_2 = x}\\ \sf{V_2=75ml}\\ \sf{V_1=200ml}\\ \sf{T_2=57°C = 330K}

Using the gas equation \sf{ \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}}

Substituting the values,

\sf{ \dfrac{70 \times 200}{300}=\dfrac{P_2 \times 75}{330}}

\sf{P_2 = \dfrac{70 \times 200 \times 330}{75 \times 300} }

\sf \: P_2 = 205.33cm \: of \: Hg

∴ The final pressure of the gas is 205.33 cm of Hg

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