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A balloon of air occupies 10.0L(V2)at25.0°C(T2)and1.00atm(P2). What temperature (T1) was it initially, if it occupied 9.40 L (V1) and was in a freezer with a pressure of 0.939 atm (P1)?

Respuesta :

Answer: The initial temperature was 263 K

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 0.939 atm

[tex]P_2[/tex] = final pressure of gas = 1.00 atm

[tex]V_1[/tex] = initial volume of gas = 9.40 L

[tex]V_2[/tex] = final volume of gas = 10.0 L

[tex]T_1[/tex] = initial temperature of gas = ?

[tex]T_2[/tex] = final temperature of gas = [tex]25^oC=273+25=298K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.939\times 9.40}{T_1}=\frac{1.00\times 10.0}{298}[/tex]

[tex]T_1=263K[/tex]

Thus the initial temperature was 263 K

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