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A helium-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.)

Respuesta :

Answer : The value of [tex]\Delta E[/tex] of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

[tex]\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}[/tex]

Molar mass of helium = 4 g/mole

[tex]\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole[/tex]

Now we have to calculate the heat.

Formula used :

[tex]q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)[/tex]

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

[tex]c_p[/tex] = specific heat of helium = 20.8 J/mol.K

[tex]T_1[/tex] = initial temperature = 310.0 K

[tex]T_2[/tex] = final temperature = 235.0 K

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

[tex]q=nc_p(T_2-T_1)[/tex]

[tex]q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K[/tex]

[tex]q=-19.5J[/tex]

Now we have top calculate the work done.

Formula used :

[tex]w=-p\Delta V\\\\w=-p(V_2-V_1)[/tex]

where,

w = work done

p = pressure of the gas = 1 atm

[tex]V_1[/tex] = initial volume = 1.21 L

[tex]V_2[/tex] = final volume = 0.99 L

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

[tex]w=-p(V_2-V_1)[/tex]

[tex]w=-(1atm)\times (0.99-1.21)L[/tex]

[tex]w=0.22L.artm=0.22\times 101.3J=22.29J[/tex]

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

Now we  have to calculate the value of [tex]\Delta E[/tex] of the gas.

[tex]\Delta E=q+w[/tex]

[tex]\Delta E=(-19.5J)+22.29J[/tex]

[tex]\Delta E=2.79J[/tex]

Therefore, the value of [tex]\Delta E[/tex] of the gas is 2.79 Joules.

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