Respuesta :
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 :
[tex]\Delta H=\Delta q\\\\\Delta q=nC_p\Delta T\\\\Q=nC_p\Delta T[/tex]
where,
Q = heat at constant pressure = ?
[tex]\Delta H[/tex] = change in enthalpy energy
n = number of moles of helium gas = 78.25 mol
[tex]\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}[/tex]
Molar mass of He = 4 g/mole
[tex]\text{Moles of }He=\frac{313g}{4g/mole}=78.25mole[/tex]
[tex]C_p[/tex] = heat capacity at constant pressure = [tex]20.8J/mol.^oC[/tex]
[tex]\Delta T[/tex] = change in temperature = [tex]-41.6^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]Q=nC_p\Delta T[/tex]
[tex]Q=(78.25mol)\times (20.8J/mol.^oC)\times (-41.6)^oC[/tex]
[tex]Q=-67708.16J=-67.7kJ[/tex]
Now we have to 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 = 1910 L
[tex]V_2[/tex] = final volume = 1643 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 (1643-1910)L[/tex]
[tex]w=267L.artm=267\times 101.3J=27047.1J=27.05kJ[/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=(-67.7kJ)+27.05kJ[/tex]
[tex]\Delta E=-40.65kJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] of the gas is -40.65 kJ.
