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Nadya [2.5K]
3 years ago
7

Consider a balloon filled with helium at the following conditions. 313 g He 1.00 atm 1910. L Molar Heat Capacity = 20.8 J/degree

C middot mol The temperature of this balloon is decreased by 41.6 degree C as the volume decreases to 1643 L with the pressure remaining constant. Determine q, w, and Delta E (in kJ) for the compression of the balloon.
Chemistry
1 answer:
yawa3891 [41]3 years ago
5 0

Answer : The value of q, w and ΔE in kilojoule are -67.7 kJ, 27.05 kJ and -40.65 kJ respectively.

Explanation :

Formula used :

\Delta H=\Delta q\\\\\Delta q=nC_p\Delta T\\\\Q=nC_p\Delta T

where,

Q = heat at constant pressure = ?  

\Delta H = change in enthalpy energy

n = number of moles of helium gas = 78.25 mol

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}

Molar mass of He = 4 g/mole

\text{Moles of }He=\frac{313g}{4g/mole}=78.25mole

C_p = heat capacity at constant pressure = 20.8J/mol.^oC

\Delta T = change in temperature = -41.6^oC

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

Q=nC_p\Delta T

Q=(78.25mol)\times (20.8J/mol.^oC)\times (-41.6)^oC

Q=-67708.16J=-67.7kJ

Now we have to calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 1910 L

V_2 = final volume = 1643 L

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

w=-p(V_2-V_1)

w=-(1atm)\times (1643-1910)L

w=267L.artm=267\times 101.3J=27047.1J=27.05kJ

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

Now we  have to calculate the value of \Delta E of the gas.

\Delta E=q+w

\Delta E=(-67.7kJ)+27.05kJ

\Delta E=-40.65kJ

Therefore, the value of \Delta E of the gas is -40.65 kJ.

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