Question

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.

Answer 1

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.