Answer :
Hi! the english version of the given question is "A helium balloon is inflated to the volume of 0.045 m3 at a temperature of 2 ° C. If the balloon is cooled to -12 ° C. What will its new volume be? Consider that the pressure does not vary". The answer is given in english.
Answer:
The new volume of balloon is [tex]0.043\;m^{3}[/tex].
Explanation:
Let's assume the helium gas inside the balloon behaves ideally.
The total number of moles of helium gas inside the balloon remains constant.
Applying combined gas law, we get:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}\;and\;P_{2}[/tex] are the initial and final pressure of the balloon respectively.
[tex]V_{1}\;and\;V_{2}[/tex] are the initial and final volume of the balloon respectively.
[tex]T_{1}\;and\;T_{2}[/tex] are the initial and final temperature in the kelvin scale respectively.
Given:
[tex]P_{1}=P_{2}[/tex]
[tex]V_{1}=0.045\;m^{3}[/tex]
[tex]T_{1}=(273+2)\;K=275\;K[/tex]
[tex]T_{2}=(273-12)\;K=261\;K[/tex]
Substituting the above values, we get:
[tex]\frac{(0.045\;m^{3})}{275\;K}=\frac{V_{2}}{261\;K}[/tex]
[tex]\Rightarrow V_{2}=0.043\;m^{3}[/tex]