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3 years ago

A sealed container at 25oC contains a gas at a pressure of 104 kPa. What is the pressure of the gas when it is heated to 225oC?

Chemistry

1 answer:

seropon [69]3 years ago

4 0

Answer:

174 kPa

Explanation:

Given that,

Initial temperature, T₁ = 25° C = 25+273 = 298 K

Final temperature, T₂ = 225°C = 225 + 273 = 498 K

Initial pressure, P₁ = 104 kPa

We need to find the new pressure. The relation between the temperature and pressure is given by :

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

So,

$P_2=\dfrac{P_1T_2}{T_1}\\\\P_2=\dfrac{104\times 498}{298}\\\\P_2=173.79\ kPa$

P₂ = 174 kPa

So, the new pressure is 174 kPa.

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Answer:

The value is $K = 8*10^{61}$

Explanation:

From the question we are told that

The equation is $2 Cr + 3Pb^{2+} \to 3Pb + 2Cr^{3+}$

The temperature is $T = 25^oC = 298 K [room \ temperature ]$

The emf at standard condition is $E^o_{cell} = 0.61 \ V$

Generally at the cathode

$3Pb^{2+}(aq) + 6 e- --> 3Pb(s)$

At the anode

$2Cr^{3+} + 6e^- \to 2Cr$

Generally for an electrochemical reaction, at room temperature the Gibbs free energy is mathematically represented as

$G = n* F * E^o_{cell}$

Here n is the no of electron with value n = 6

F is the Faraday's constant with value 96487 J/V

=> $G = 6 * 96487 * 0.61$

=> $G = 3.5 *10^{5} \ J$

This Gibbs free energy can also be represented mathematically as

$G = RTlogK$

Here R is the cell constant with value 8.314J/K

K is the equilibrium constant

From above

=> $K = antilog^{\frac{G}{ RT} }$

Generally antilog = 2.718

=> $K = 2.718^{\frac{3.5 *10^5}{ 8.314* 298} }$

=> $K = 8*10^{61}$

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The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

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The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$