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:

$PV=nRT\\\\PV=\frac{w}{M}RT$

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:

$(3.646atm)\times (88.84L)=\frac{609.64g}{38g/mole}\times (0.0821L.atm/mole.K)\times (T)$

$T=245.9K$

Therefore, the temperature of the gas is, 245.9 K