Answer:
0.25 L
Explanation:
[tex]P_1[/tex] = Initial pressure = 1 atm
[tex]T_1[/tex] = Initial Temperature = 20 °C
[tex]V_1[/tex] = Initial volume = 4.91 L
[tex]P_2[/tex] = Final pressure = 5.2 atm
[tex]T_2[/tex] = Final Temperature = -196 °C
[tex]V_2[/tex]= Final volume
From ideal gas law we have
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L[/tex]
The pressure experienced by the balloon is 0.25 L
