Question

It is found that, when a dilute gas expands quasistatically from 0.80 to 3.0 l, it does 280 j of work. Assuming that the gas temperature remains constant at 300 k. How many moles of gas are present?

Answer 1

The number of moles of gas present is 0.0850 mol

Given

initial volume of the gas $v_{i}$= 0.80 L

final volume of the gas $v_{f}$ = 3.0L

work done by the gas, W = 280 J

constant temperature, T = 300 K

gas constant, R = 8.31 J/mol.K

work done by gas at constant temperature is given as;

$W= \int\limits^a_b {P} \, dv$

where a =final volume of the gas

and b = initial volume of the gas

W=nRT*ln$(\frac{v_{f} }{v_{i} } )$

$n=\frac{W}{RT*ln\frac{v_{f} }{v_{i} } }$

$n=\frac{280}{8.31*300*ln\frac{3}{0.80 } }$

n=0.0850 mol

Therefore, the number of moles of gas present is 0.0850 mol

To learn more about mole concept

brainly.com/question/17010807

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