Question

What is the temperature of a gas if the container has a volume of 2,300 mL, with a pressure of 932 mmHg and 3.51 moles?

Answer 1

Answer:

The preceding temperature is equivalent to approximately 9.6 K => -263.5 °C

Explanation:

Given the provided problem, the formula may be substituted by the corresponding values:

Pressure (P)= 932 mmHg ==> 1.2 atm

Volume (V) = 2,300 mL ==> 2.31 L

Moles (n) = 3.51 moles ==> 3.51 mol

Proportional constant of ideal gas constant: (R) = 0.08206

Therefore, we can equate the following:

T= (PV)

     nR

T= (1.2 atm · 2.31 L)

   (3.51 mol · 0.08206)

T= 9.6 K ==> -263.5 °C

Thus, the cumulative temperature given the substituents/substance is equivalent to -263.5 °C.