The mass of a hot-air balloon and its cargo ( but not the hotair inside) is 250

Question

The mass of a hot-air balloon and its cargo ( but not the hotair inside) is 250 kg. The air outside is 15 degrees C and ata pressure of 101 kPa. If the volume of the balloon is 600 meterscubed, what must the temperature of the hot air in the balloon bein order for it to just barely lift off (neutralbuoyancy)? Take the density of air to be 1.20 kg/meterscubed?? The answer should be around 168 degrees C Thanks. The mass of a hot-air balloon and its cargo ( but not the hotair inside) is 250 kg. The air outside is 15 degrees C and ata pressure of 101 kPa. If the volume of the balloon is 600 meterscubed, what must the temperature of the hot air in the balloon bein order for it to just barely lift off (neutralbuoyancy)? Take the density of air to be 1.20 kg/meterscubed?? The answer should be around 168 degrees C Thanks.

Explanation / Answer

m1 =1V1 m2 =2V2 where is the density of thegas in the balloon. m1/m2 =1/2 and m1-m2 =(1-2/1)m1 = 250kg P1V1 =N1RT1   P2V2 = N2RT2 N1/N2= T2/T1 since pressure and volume of gas in balloon is constant Also, since N = K (number of moles proportionalto density) 1/2=T2/T1 (1-T1/T2)m1 = 250 andm1 = 600 * 1.2 = 720 kg (1-T1/T2)*720 =250   T1/T2 =(720-250)/720 = .653 Since T1 = 288   T2 = 441 degK = 168 deg C m1/m2 =1/2 and m1-m2 =(1-2/1)m1 = 250kg P1V1 =N1RT1   P2V2 = N2RT2 N1/N2= T2/T1 since pressure and volume of gas in balloon is constant Also, since N = K (number of moles proportionalto density) 1/2=T2/T1 (1-T1/T2)m1 = 250 andm1 = 600 * 1.2 = 720 kg (1-T1/T2)*720 =250   T1/T2 =(720-250)/720 = .653 Since T1 = 288   T2 = 441 degK = 168 deg C P1V1 =N1RT1   P2V2 = N2RT2 N1/N2= T2/T1 since pressure and volume of gas in balloon is constant Also, since N = K (number of moles proportionalto density) 1/2=T2/T1 (1-T1/T2)m1 = 250 andm1 = 600 * 1.2 = 720 kg (1-T1/T2)*720 =250   T1/T2 =(720-250)/720 = .653 Since T1 = 288   T2 = 441 degK = 168 deg C (1-T1/T2)m1 = 250 andm1 = 600 * 1.2 = 720 kg (1-T1/T2)*720 =250   T1/T2 =(720-250)/720 = .653 Since T1 = 288   T2 = 441 degK = 168 deg C

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