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Marina CMI [18]
3 years ago
12

A gas occupies a 1.5 L container at 25 degrees Celsius and 2.0 atm. If the gas is transferred to a 3.0 L container at the same t

emperature, what will be its new pressure?
a) 3.0 atm
b) 2.0 atm
c) 1.0 atm
d) 4.0 atm
Chemistry
2 answers:
dybincka [34]3 years ago
5 0

Answer:

Option c. 1 atm

Explanation:

We can do this exercise in two ways.

1. using the ideal gas equation which is PV = nRT

2. Using boyle's law.

Now, with the first option, would be longer, because we will need to calculate the number of moles present in the first container, and then, replace it in the second container to calculate P.

However, in this case, is easier to use Boyle's law.

Boyle's law is a relation concerning the compression and expansion of a gas at constant temperature.

In this case, we can see that the gas, is being expanded to a major volume, but temperature is constant (same 25 °C), therefore, we can use the following expression to calculate the new pressure:

P1*V1 = P2*V2

Now, let's replace the data provided in this expression, and solve for P2:

P2 = P1*V1 / V2

P2 = 2 * 1.5 / 3

P2 = 3 / 3

P2 = 1 atm

This would be the new pressure of the gas at constant temperature

harkovskaia [24]3 years ago
4 0

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1 x V1 / V2

P2 = 2.0 x 1.5 / 3

<span>P2 = 1 atm</span>

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Answer:

∆H°rxn in kJ/mol AgCl = - 59.61 kJ/mol

Explanation:

To solve this question we need to calculate the heat absorbed by the cup calorimeter and by the water in the solutions. We will also need to calculate the amount in moles produced by the reaction since we want to know the ∆H°rxn in kJ/mol AgCl .

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qw = 200 g x 4.18 J/gºC x 0.7 ºC = 585.20 J

Heat Absorbed by the calorimeter :

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We have to change the sign to this quantity since it is an exotermic reaction  ( ΔT is positive ) and have the ∆Hrxn

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but this  is not what we are being asked since this heat was released by the formation of  0.0100 mol of AgCl so finally  

∆H°rxn = -596.05 J /0.01 mol  = -59,605 J/mol x 1 kJ/1000J = -59.61 kJ/mol

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