Question

a) volumul ocupat la 227°C şi 5 atm de on
528 g de CO2
Urgent!!!​

Answer 1

PV = nRT ; n = m/M

PVM = mRT

P= 5 atm, m= 528 g M.CO2 = 12+2x16 = 44 g/mol

T= 273+227 = 500 K

=> V = mRT/PM

= 528x0,082x500/5x44 = 98,4 L

Answer 2

PV = $\pi$RT

n=m/M

PVM = mRT

P - 5 atm

m = 528 g M CO2 = 12+2×16=12+32=44$\frac{g}{mol}$

T = 273K+227K=500K

528×0,082 ×500/5 × 44} = 98.4