Answer:
The temperature of the sample is -128.14°C .
Explanation:
1st) It is necessary to convert the1.35g of He gas to moles, using the molar mass of helium (4g/mol):
[tex]1.35g*\frac{1mol}{4g}=0.3375moles[/tex]
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:
[tex]\begin{gathered} P*V=n*R*T \\ 0.127atm*31.6L=0.3375mol*0.082\frac{atm*L}{mol*K}*T \\ 4.0132atm*L=0.027675\frac{atm*L}{K}*T \\ \frac{4.0132atm*L}{0.027675\frac{atmL}{K}}=T \\ 145.01K=T \end{gathered}[/tex]
3rd) Finally, we can convert the temperature from K to °C:
[tex]145.01-273.15=-128.14°C[/tex]
So, the temperature of the sample is -128.14°C.
