Question

On a hot summer day, the density of air at atmospheric pressure at 32.5°C is 1.1242 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure? mol (b) Avogadro's number of air molecules has a mass of 2.86 x 10-2 kg. What is the mass of 1.00 m3 of air? (Assume air is an ideal gas.) kg (c) Does the value calculated in part (b) agree with the stated density of air at this temperature? (Consider that it does if the values are within 10% of each other.) O Yes Ο Νο

Answer 1

ρ = 1.1242 kg/m^3

T= 32.5 = 32.5+273 -= 305.5 k

P = 1 atm = 1*1.013*10⁵ pa

V = 1 m³

n = PV/RT = 1*1.013*10^5*1 /( 8.3114*305.5) = 39.8955

number of mole =  39.8955

B )

Avogadro no of molecule means one mole so the mass of mole air = 2.86*10^-2 kg so totla mass of air =

=  39.8955*2.86*10^-2 =

= 1.1410113 kg answer

hence mass of 1 m³ is =1.1410113

C)

Yes

1.1410113 and  1.1242 are comparable with the given limit