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

Step-by-step explanation:

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

The combined gas equation is,

`(P_1V_1)/(T_1)=(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:

`(0.939* 9.40)/(T_1)=(1.00* 10.0)/(298)`

`T_1=263K`

Thus the initial temperature was 263 K