Answer:
The temperature of the sample is -128.14°C .
Step-by-step explanation:
1st) It is necessary to convert the1.35g of He gas to moles, using the molar mass of helium (4g/mol):
![1.35g*(1mol)/(4g)=0.3375moles](https://img.qammunity.org/2023/formulas/chemistry/college/h9fs06q4rvh11fnrtgnxd9hf4mqcnvrk5b.png)
Now we know that are 0.3375moles in 1.35g of He.
2nd) With the Ideal gases formula, we ca replace the values of pressure, volume and moles to find the temperature:
![\begin{gathered} P*V=n*R*T \\ 0.127atm*31.6L=0.3375mol*0.082(atm*L)/(mol*K)*T \\ 4.0132atm*L=0.027675(atm*L)/(K)*T \\ (4.0132atm*L)/(0.027675(atmL)/(K))=T \\ 145.01K=T \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/wrsrp6twry1bfx7lpk2nouelttgijm5s48.png)
3rd) Finally, we can convert the temperature from K to °C:
![145.01-273.15=-128.14°C](https://img.qammunity.org/2023/formulas/chemistry/college/j8133yjd8saw3k8zdrx9b7zmropdxa7rsx.png)
So, the temperature of the sample is -128.14°C.