A5.00 liter gas sample is collected at a temperature and pressure of 27.0 degree c and 1.20 atm. it is desired to transfer the gas to a 3.00 liter container at a pressure of 1.00 atm. what must be the temperature of the gas in the 3.00 liter container?
a) 417 k
b) 216 k
c) 13.5 degree c
d) 150 k

The temperature of the gas in the 3.00 liter container, must be 150K

Explanation:

Let's apply the Ideal Gas Law, to find out the moles

P . V = n . R . T

1,20 atm . 5L = n . 0,082 L.atm/mol.K . 300 K

(1,20 atm . 5L) /  (0,082 mol.K/L.atm . 300 K) = n

6/24,6 mol = n = 0,244 moles

We have the moles now, so let's find the temperature in our new conditions.

P . V = n . R . T

1 atm . 3L = 0,244 moles . 0,082 L.atm/mol.K . T° in K

(1 atm . 3L / 0,244 moles . 0,082 mol.K/L.atm) = T° in K

3/20,008 K = T° in K = 150K

the answer for sodium is 0.186

the answer for rubidium is 0.251