Question

The pressure of a 609.64 gram sample of F2 in a 88.84 L container is measured to be 2770.96 torr. What is the temperature of this gas in Kelvin?

Answer 1

Answer : The temperature of the gas is, 245.9 K

Explanation :

To calculate the temperature of gas we are using ideal gas equation:

`PV=nRT\\\\PV=(w)/(M)RT`

where,

P = pressure of gas = 2770.96 torr = 3.646 atm

Conversion used : (1 atm = 760 torr)

V = volume of gas = 88.84 L

T = temperature of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas = 609.64 g

M = molar mass of
`F_2` gas = 38 g/mole

Now put all the given values in the ideal gas equation, we get:

`(3.646atm)* (88.84L)=(609.64g)/(38g/mole)* (0.0821L.atm/mole.K)* (T)`

`T=245.9K`

Therefore, the temperature of the gas is, 245.9 K