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Using 0.082 as your R value, what is the temperature of a 100L container of gas containing 64 mols at 1.75 atm?

Sagot :

To solve this problem we must assume that the gas behaves like an ideal gas, that is, that the gas molecules do not interact with each other. Now, to find the temperature we apply the gas law which tells us:

[tex]PV=nRT[/tex]

Where,

P is the pressure of the gas = 1.75atm

V is the volume of the gas = 100L

n is the number of moles =64moles

R is a constant = 0.082 (atm.L)/(mol.K)

T is the temperature of the gas =?

Now, we must clear the temperature of the equation and replace the values that we know:

[tex]\begin{gathered} T=\frac{PV}{nR} \\ T=\frac{1.75\text{atm}\times100L}{64mol\times0.082\frac{atm.L}{mol.K}}=33.35K \end{gathered}[/tex]

So, the temperature of the gas is 33.35 K