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.

your answer would be a. 12.5 kg

the answer is true. hope it .