Question

A gas is contained in a thick-walledballoon. When the pressure changesfrom 1.21 atm to 2.52 atm, thevolume changes from 3.75 L to 1.72L and the temperature changes from293 K toK.

Answer 1

280 is the correct answer on acellus

Step-by-step explanation:

Answer 2

This is an ideal gas problem. We can use the following formula (combined gas law):

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2),`

where P represents pressure, V volume, and T temperature. Subindex 1 indicates the initial state and subindex 2 indicates the final state.

We have the following data:

`\begin{gathered} P_1=1.21\text{ atm} \\ P_2=2.52\text{ atm} \\ V_1=3.75\text{ L} \\ V_2=1.72\text{ L} \\ T_1=293\text{ K} \\ T_2=? \end{gathered}`

So, we have to clear the final temperature (T2) in the formula of combined gas law and we're going to obtain:

`T_2=(P_2V_2T_1)/(P_1V_1)`

The final step is to replace all the given data:

`\begin{gathered} T_2=\frac{2.52\text{ atm}\cdot1.72L\cdot293K}{1.21\text{ atm}\cdot3.75L}, \\ T_2=279.885\text{ K}\approx280K. \end{gathered}`

The answer is that the final temperature would be 280 K.