Ahelium-filled balloon at 310.0 k and 1 atm, contains 0.05 g he, and has a volume of 1.21 l. it is placed in a freezer (t = 235.0 k), and its volume decreases to 0.99 l. find δe for the gas (in joules). (cp of he = 20.8 j/mol k.)

Answer : The value of of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

Molar mass of helium = 4 g/mole

Now we have to calculate the heat.

Formula used :

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

= specific heat of helium = 20.8 J/mol.K

= initial temperature = 310.0 K

= final temperature = 235.0 K

Now put all the given values in the above formula, we get:

Now we have top calculate the work done.

Formula used :

where,

w = work done

p = pressure of the gas = 1 atm

= initial volume = 1.21 L

= final volume = 0.99 L

Now put all the given values in the above formula, we get:

conversion used : (1 L.atm = 101.3 J)

Now we  have to calculate the value of of the gas.

Therefore, the value of of the gas is 2.79 Joules.

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.