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nignag [31]
3 years ago
7

An ideal gas contained in a piston which is compressed. The gas is insulated so that no heat flows into or out of it. 1) What ha

ppens to the temperature of the gas when it is compressed?
Chemistry
1 answer:
Goryan [66]3 years ago
5 0

Answer:Temperature increases

Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.

We know ideal gas equation to be PV=nRT where

   P=pressure

   V=Volume

   T=Temperature

   R=Real gas constant

   n=Number of moles

since the gas is insulated such that no heat goes into or out of the system .

When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.

Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.

From the equation PV=nRT

Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.

As the ideal  gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and  so the pressure would increase .

The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.

So here we can say that work done on the system by surroundings leads to increase in temperature of the system.

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Answer : The value of \Delta H for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)     \Delta H_1=-571.6kJ/mole

Now we have to determine the value of \Delta H for the following reaction i.e,

2H_2O(l)\rightarrow 2H_2(g)+O_2(g)    \Delta H_2=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H_2 for the reaction will be:

\Delta H_2=-(-571.6kJ/mole)

\Delta H_2=+571.6kJ/mole

Hence, the value of \Delta H for the reaction is +571.6 kJ/mole.

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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

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The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

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By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

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