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Lena [83]
3 years ago
12

A tire at 21°C has a pressure of 0.82 atm. Its temperature decreases to –3.5°C. If there is no volume change in the tire, what i

s the pressure after the temperature change?
Chemistry
2 answers:
Alenkinab [10]3 years ago
8 0

Answer : The pressure after the temperature change is, 0.75 atm.

Solution :

According to the Gay-Lussac's Law, the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T     (At constant volume and number of moles)

Or,

\frac{P_1}{P_2}=\frac{T_1}{T_2}

where,

P_1 = initial pressure = 0.82 atm

P_2 = final pressure = ?

T_1 = initial temperature = 21^oC=273+21=294K

T_2 = final temperature = -3.5^oC=273+(-3.5)=269.5K

Now put all the given values in the above formula, we get the final pressure of the gas.

\frac{0.82atm}{P_2}=\frac{294K}{269.5K}

P_2=0.75atm

Therefore, the pressure after the temperature change is, 0.75 atm

Damm [24]3 years ago
4 0

The correct answer is A 0.75 atm


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Answer:

The final temperature of the mixture is 28.11 °C

Explanation:

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enthalpy change is – 26 kJ per mol BaSO4

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Step 2: The balanced equation

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Step 4: Calculate mass

Mass = volume * density

Mass = 2000 mL * 1g/mL

Mass = 2000 grams

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For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol BaSO4

There is no limiting reactant, both Ba(NO3)2 and Na2SO4 will be completely be consumed (1 mol). We'll have 1.0 mol of BaSO4 produced.

Step 6: Calculate Q

Q = - ΔH

ΔH is negative so the reaction is exothermic, what means the temperature increases

Q is always positive, so Q = 26kJ = 26000 J

Step 6: Calculate the heat transfer

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⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m =the mass of the solution = 2000 grams

⇒with c= the specific heat of the solution = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = T2 - 25.0

26000 = 2000 * 4.18 * (T2 - 25.0 °C)

3.11 = T2 - 25.0 °C

T2 = 25.0 + 3.11 °C

T2 = 28.11 °C

The final temperature of the mixture is 28.11 °C

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