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Anton [14]
3 years ago
7

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)?
Chemistry
1 answer:
andrey2020 [161]3 years ago
4 0

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,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.939 atm

P_2 = final pressure of gas = 1.00 atm

V_1 = initial volume of gas = 9.40 L

V_2 = final volume of gas = 10.0 L

T_1 = initial temperature of gas = ?

T_2 = final temperature of gas = 25^oC=273+25=298K

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

\frac{0.939\times 9.40}{T_1}=\frac{1.00\times 10.0}{298}

T_1=263K

Thus the initial temperature was 263 K

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