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BaLLatris [955]
4 years ago
8

(14.1.50) 0.03 mol of helium are enclosed in a cylinder with a piston which maintains a constant pressure in the helium. Initial

ly the helium is immersed in thermal bath with at 250K. Suddenly, the helium is removed from this thermal bath and plunged into a thermal bath at temperature 350K. Determine (a) the increase in the entropy of the helium. (Hint the entropy change for the helium is the same as if it had been taken through this temperature change by a reversible process.) (b) the increase in entropy of the universe. (Answer: 0.032 J/K)

Chemistry
1 answer:
icang [17]4 years ago
3 0

Answer:

a) 0.210 j /k

b) 0.032 j/k

Explanation:

Find the attachment for solution

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A force of 25 newtons is exerted over an area of 50 square meters. Calculate
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Answer:

the pressure exerted in pascals is 0.5 Pa

Explanation:

The computation of the pressure exerted in pascals is shown below:

As we know that

Pressure = force ÷ area

= 25 ÷ 50

= 0.5 Pa

Hence, the pressure exerted in pascals is 0.5 Pa

We simply applied the above formula so that the correct pressure could come

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What does combustion mean
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HNO3 + KOH + H2O + KNO3<br><br> ACID:<br> BASE:<br> SALT:
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Explanation:

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base, KOH

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What is one source of CO?
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Explanation:

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5 0
4 years ago
Assume you mix 100.0 mL of 200 M CsOH with 50.0 mL 0f 0.400 M HCl in a coffee cup calorimeter. A reaction occurs. The temperatur
Varvara68 [4.7K]

Answer:

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

Explanation:

Step 1: Data given

Volume of a CsOH solution = 100.0 mL

Molarity of a CsOH solution = 0.200 M

Volume of HCl solution = 50.0 mL

Molarity of HCl solution = 0.400M

The temperature of both solutions before mixing was 22.50 °C, and it rises to 24.28 °C after the acid-base reaction.

Density = 1.00 g/mL

Specific heat = 4.2 J/gK = 4.2 J/g°C

Step 2: The balanced equation

CsOH + HCl → CsCl + H2O

Step 3: Calculate the energy

Q = m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass of the solution = (100+ 50 mL) * 1.00 g/mL = 150 grams

⇒with c = the specific heat of the solution = 4.2 J/g°C

⇒with ΔT = The change of temperature = T2 - T1  = 24.28 °C - 22.50 °C = 1.78 °C

Q = 150 grams * 4.2 J/g°C * 1.78 °C

Q = 1121.4 J

Step 4: Calculate moles CsOH

Moles CsOH = molarity CsOH * volume CsOH

Moles CsOH = 0.200M * 0.100 L

Moles CsOH = 0.0200 moles

Step 5: Calculate  the enthalpy change for the reaction per mole of CsOH

ΔH is negative since this is an exothermic reaction

ΔH = -Q/moles

ΔH = -1121.4 J / 0.0200 moles

ΔH = -56070 J/mol = -56.1 kJ/mol

The enthalpy change for the reaction per mole of CsOH is -56.1 kJ/mol

4 0
4 years ago
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