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sergeinik [125]
3 years ago
9

In an isothermal process, 1.59 moles of an ideal gas is compressed to one-fifth of its initial volume at 285 K. What quantity of

heat is added to, or removed from, the system during this process
Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Answer:

−6063.54 J  of heat is removed from the system during this process.

Explanation:

Given that :

in an isothermal process

number of moles of an ideal gas =  1.59 moles

which is compressed to one- fifth of its initial volume

Let say the initial volume = x

the final volume will be = x/5

Temperature = 285 K

The objective is to determine what quantity of heat is added to, or removed from, the system during this process.

In an isothermal process;

q = w

where ;

w = nRT In (V₂/V₁)

w = 1.59 × 8.314 × 285 × In (x/5 /x)

w = 1.59 × 8.314 × 285 × In (1/5)

w = −6063.54 J

Thus;  −6063.54 J  of heat is removed from the system during this process.

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Answer: 2.00 atm

Explanation:

P1/T1 = P2/T2

1.3317/273 = P2/410

P2 = 2.00 atm

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49.Which of the following statements regarding chemical equilibrium is true?
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A 100.00 ml of volume of 0.500 M HCl was mixed with 100.00 ml of 0.500 M KOH in a constant pressure calorimeter. The initial tem
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Answer:

The heat of the reaction = -1985 J = -1.985 kJ

The enthalpy of the reaction is -39.7 kJ/ mol

Explanation:

<u>Step 1: </u>Data given

Volume of HCl = 100 mL the heat of the reaction = 0.1 L

Molarity of HCl solution = 0.500 M

Volume of KOH = 100 mL = 0.1 L

Molarity of KOH solution = 0.500 M

Initial temperature = 23.0 °C

Final temperature = 25.5 °C

Specific heat of the solution = 3.97 J/°C *g

Density of the solution = 1g/ mL

<u>Step 2: </u>Calculate heat

Q = m*c*ΔT

with m = the mass of both solution : 100g + 100 g ( since density = 1g/mL) = 200 g

c = the specific heat = 3.97 J/°C*g

ΔT  = T2 -T1 = 25.5 = 23 = 2.5 °C

Q = 200g *3.97 J/°C*g * 2.5°C = 1985 J  (= -1985 J because it's exothermic)

<u />

<u>Step 3:</u> Calculate the number of moles

Number of moles = Molarity * Volume

Number of moles = 0.5 * 0.1 L = 0.05 moles

(Moles of the acid are equal to the moles of water produced.

Moles of solution = 0.05 moles)

<u>Step 4: </u>Calculate the enthalpy of the reaction

ΔH = heat change /Number of moles

    = -1.985 kJ/ 0.05 moles

   =- 39700 J/mol = -39.7 kJ/ mol

The enthalpy of the reaction is -39.7 kJ/ mol

The enthalpy of the reaction is negative, this means the reaction is exothermic ( which means the final temperature is higher than the initial temperature.)

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