Answer: The volume of balloon at a height of 20 km is
![1.32* 10^5L](https://img.qammunity.org/2020/formulas/chemistry/college/3yrniv6or73wlkbfyirn6xkmcfmpfv97qn.png)
Step-by-step explanation:
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
![(P_1V_1)/(T_1)=(P_2V_2)/(T_2)](https://img.qammunity.org/2020/formulas/chemistry/high-school/50mvdq8vszyhvl7dxwm291qdlkdomofz2f.png)
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
![P_1=745torr\\V_1=1.41* 10^4L\\T_1=21^oC=[21+273]K=294K\\P_2=63.1torr\\V_2=?L\\T_2=-40^oC=[-40+273]K=233K](https://img.qammunity.org/2020/formulas/chemistry/college/wdcfzot9ve0x1eda2fzknvs6d2cmov64bf.png)
Putting values in above equation, we get:
![(745torr* 1.41* 10^4L)/(294K)=(63.1torr* V_2)/(233K)\\\\V_2=(745* 1.41* 233)/(294* 63.1)=1.32* 10^5L](https://img.qammunity.org/2020/formulas/chemistry/college/tm2jdzztazij8g44t0nhsycjidtx1mhcnp.png)
Hence, the volume of balloon at a height of 20 km is
![1.32* 10^5L](https://img.qammunity.org/2020/formulas/chemistry/college/3yrniv6or73wlkbfyirn6xkmcfmpfv97qn.png)