A hot air balloon consists of a very large bag filled with heated air at 120.0 °

Question

A hot air balloon consists of a very large bag filled with heated air at 120.0 °C produced from

a propane burner. It has a basket hanging under it in which several people can ride. The average

hot air balloon has a volume of 2.80 × 10^3m3 when fully inflated.

a)Determine the work (in kJ) needed to inflate the balloon under an external air pressure of 1.00atm. (1 mark)

b)The specific heat of the air found in the balloon has a value of 0.9313 J g-1°C-1. The hot air which contains some carbon dioxide and water vapour, products of the propane combustion, has an average molar mass of 32.0 g mol-1 at a pressure of 1.00 atm. If the initial temperature of the air was 25.0 °C, how much heat (in kJ) is required to warm the air up to 120.0 °C?

Explanation / Answer

(a) Amount of work done

= -PV

= - 1 atm * ( 2.80*103 - 0)m3

= - 1 atm * (2.80*103 m3)

= - 1 atm * (2.80*103 * 1000)L

= - 28*105 L.atm

= (- 28*105 L.atm) ( 101.3 J / 1L.atm)

= - 283640 kJ

(b) specific heat of the air found in the balloon = 0.9313 J g-1°C-1

molar mass of air inside balloon = 32.0 g mol-1

We need to find the mass of air inside balloon. To do that first we require to calculate the density of air. We can do that from the ideal gas law, PV = nRT

We can write, PV = (m/M) RT       ;where m = mass of air, M = molar mass of air

or, PM = (m/V)RT = RT                 ;where = density

or, = PM/RT

= (1 atm)( 32.0 g mol-1) / (0.082 L.atm.mol-1.K-1)(393 K)

= 0.99 g/L

Amount of air inside the balloon

= (0.99 g/L)( 2.80*103 * 1000 L)

= 2780363 g

Heat required, Q = m.Cp.dT

= (2780363 g)( 0.9313 J g-1°C-1)(393 - 298)K

= 245988506 J

= 245988.51 kJ

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