A gas stream containing 45wt% butane and 55wt% pentane is subjected to heating from 180K to 260K at constant pressure. calculate the required heat input per kg of this mixture?

neglecting changes in kinetic and potential energies.

1 Answer
Jun 11, 2018

Answer:

The required heat input is 184 kJ/kg.

Explanation:

Assume that we have 1 kg of the mixture.

Then we have 450 g of butane and 550 g of pentane.

Step 1. Calculate the heat input for the butane

Butane is a liquid between 180 K and 260 K.

The formula for the quantity of heat #q# transferred is

#color(blue)(bar(ul(|color(white)(a/a)q = mC_text(s)ΔTcolor(white)(a/a)|)))" "#

where

#m color(white)(ll) =# the mass of the object
#C_text(s) color(white)(l) =# its specific heat capacity
#ΔT =# its change in temperature

In this problem,

#m color(white)(m)= "450 g"#
#C_text(s) color(white)(ll)= "132.42 J·K"^"-1""mol"^"-1"#
#ΔT = T_text(f) - T_text(i) = "260 K - 180 K = 80 K"#

The first task is to convert the mass of the butane to moles.

#n = 450 color(red)(cancel(color(black)("g"))) × "1 mol"/(58.12 color(red)(cancel(color(black)("g")))) = "7.743 mol"#

Now we can determine the quantity of heat needed.

#q_text((butane) = 7.743 color(red)(cancel(color(black)("mol"))) × "132.42 J"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 80 color(red)(cancel(color(black)("K"))) = 8.20 × 10^4color(white)(l)"J"#

Step 2. Calculate the heat input for the pentane

Pentane is a liquid between 180 K and 260 K.

Here,

#m color(white)(m)= "550 g"#
#C_text(s) color(white)(ll)= "167.19 J·K"^"-1""mol"^"-1"#

We must convert the mass of the pentane to moles.

#n = 550 color(red)(cancel(color(black)("g"))) × "1 mol"/(72.15 color(red)(cancel(color(black)("g")))) = "7.623 mol"#

Now we can determine the quantity of heat needed.

#q_text(pentane) = 7.623 color(red)(cancel(color(black)("mol"))) × "167.19 J"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 80 color(red)(cancel(color(black)("K"))) = 1.02 × 10^5color(white)(l)"J"#

Step 3. Calculate the total heat required

#q_text(total) = q_text(butane) + q_text(pentane) = (8.20 × 10^4 + 1.02 × 10^5)color(white)(l)"J" = 1.84 × 10^5 color(white)(l)"J" = "184 kJ"#