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Alexandra [31]
2 years ago
13

An aerosol can with a temperature of 292 K and a pressure of 1.25 atm was left outside in the sun on a hot summer day. The press

ure inside the can rose to 1.33 atm. What was the resulting temperature in the can?
322K
1,600K
311K
274K

ik its not A
Chemistry
1 answer:
garik1379 [7]2 years ago
6 0

Answer:

310.69K

Explanation:

Given parameters:

Initial temperature T₁ = 292K

Initial pressure P₁ = 1.25atm

Final pressure P₂ = 1.33atm

Unknown:

Final temperature T₂ = ?

Solution

To find the unkown, we need to apply the combined gas law. From the combined gas law, it can be deduced that at constant volume, the pressure of a give mass or mole of gas varies directly with the absolute temperature.

Since the same aerosol can is heated, the volume is constant.

          \frac{P_{1} }{T_{1} } =  \frac{P_{2} }{T_{2} }

Now, we have to make T₂ the subject of the formula:

      T₂ =  \frac{P_{2}  x T_{1}  }{P_{1} }

       T₂ =  \frac{1.33  x 292  }{1.25 } =  310.69K

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<u>Answer:</u> The heat required for the process is 4.24 kJ

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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Given mass of benzene = 24.8 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

\text{Moles of benzene}=\frac{24.8g}{78.11g/mol}=0.318mol

To calculate the enthalpy change of the reaction, we use the equation:

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where,

q = amount of heat absorbed = ?

n = number of moles = 0.318 moles

\Delta H_{rxn} = enthalpy change of the reaction  = 30.7 kJ/mol

Putting values in above equation, we get:

30.7kJ/mol=\frac{q}{0.318mol}\\\\q=(30.7kJ/mol\times 0.318mol)=4.24kJ

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