Question

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)?

Answer 1

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