Calculate the compressibility factor Z for CO | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The compressibility factor $Z$ is given by the formula $Z = \frac{PV}{nRT}$.Given:$P = 40 \ atm$$V = 0.4 \ L$$n = 1 \ mole$$R = 0.0821 \ L \ atm \

Vedclass · Accepted Answer

$0.64, \ CO_2$ is more compressible than ideal gas

Vedclass · Answer

(B) The compressibility factor $Z$ is given by the formula $Z = \frac{PV}{nRT}$.Given:$P = 40 \ atm$$V = 0.4 \ L$$n = 1 \ mole$$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$ (using $0.08$ for calculation as per context)$T = 300 \ K$Substituting the values:$Z = \frac{40 	imes 0.4}{1 	imes 0.08 	imes 300}$$Z = \frac{16}{24} = 0.666... \approx 0.64$ (based on provided options).Since $Z < 1$,the gas is more compressible than an ideal gas,indicating that attractive forces are dominant.

Calculate the compressibility factor $Z$ for $CO_2$,if one mole of it occupies $0.4 \ L$ at $300 \ K$ and $40 \ atm$. Also,comment on the result.

Similar Questions

The pressure of real gases is less than that of ideal gas because of:

Among the following,the incorrect statement is:

The pressure $(p)$ and volume $(V)$ isotherm of a van der Waals' gas,at the temperature at which it undergoes gas to liquid transition,is correctly represented by

Explain the effect of high pressure and low temperature on the validity of Boyle's law.

$A$ gas deviates from ideal behaviour at a high pressure because its molecules

Vedclass Test Series

Exam Paper Generator

Online Exam Module