Question

(a) Calculate the P and T values for which H2(g) is in a corresponding state to CH (g) at 10.0 bar and 310 K. The critical constants are the following: P.(H2)- 12.93 bar, T(H2) 32.98 K P (CH) 45.99 bar, T(CH4) 190.56 K. (b) Considering how the critical constants T and P, are related to the van der Waal's constants a and b (see formulas on back page), which gas do you conclude has stronger intermolecular attractive forces?

Answer 1

vanderwaal's constant 'a' is relatable to intermolecular force of attraction. Higher the value of 'a' greater is the intermolecular force. Highest value of 'a' is 96.5 bar.L².mol^-2 for C₁₅H₃₂

T, 190-56 lo 5.32 ニ2-811 bar