The pressure exerted by 5 mol of a real gas | vedclass

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

Question

(B) The van der Waals equation for a real gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$ Given: $n = 5 \ mol$,$V = 1 \ L$,$T = 47 + 273 = 320 \ K$,$a

Vedclass · Accepted Answer

$77.2$

Vedclass · Answer

(B) The van der Waals equation for a real gas is: $(P + \frac{an^2}{V^2})(V - nb) = nRT$ Given: $n = 5 \ mol$,$V = 1 \ L$,$T = 47 + 273 = 320 \ K$,$a = 3.592 \ atm \ L^2 \ mol^{-2}$,$b = 0.0427 \ L \ mol^{-1}$,$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$ Substituting the values: $(P + \frac{3.592 	imes 5^2}{1^2})(1 - 5 	imes 0.0427) = 5 	imes 0.0821 	imes 320$ $(P + 89.8)(1 - 0.2135) = 131.36$ $(P + 89.8)(0.7865) = 131.36$ $P + 89.8 = \frac{131.36}{0.7865} \approx 167.02$ $P = 167.02 - 89.8 = 77.22 \ atm$ Thus,the pressure is approximately $77.2 \ atm$.

The pressure exerted by $5 \ mol$ of a real gas in a $1 \ L$ container at $47 \ ^oC$ is ......... $atm$. (Given: $a = 3.592 \ atm \ L^2 \ mol^{-2}$,$b = 0.0427 \ L \ mol^{-1}$)

Similar Questions

If the compressibility factor $(Z)$ of a real gas is $1.05$ at $STP$,what is the molar volume of the real gas (in $dm^3$)?

The cooling in a refrigerator is due to

Dominance of strong repulsive forces among the molecules of the gas ($Z =$ compressibility factor)

Assertion : Compressibility factor for hydrogen varies with pressure with positive slope at all pressures.
Reason : Even at low pressure,repulsive forces dominate hydrogen gas.

Which statement is correct regarding a real gas?

Vedclass Test Series

Exam Paper Generator

Online Exam Module