Answer:
The number of moles of gas present is 0.0629 mol.
Explanation:
Given;
initial volume of the gas, [tex]v_i[/tex] = 0.4 L = 0.4 x 10
final volume of the gas, [tex]v_f[/tex] = 3.5 L
work done by the gas, W = 340 J
constant temperature, T = 300 K
gas constant, R = 8.31 J/mol.K
work done by gas at constant temperature is given as;
[tex]W = \int\limits^{v_f}_{v_i} {Pdv} \\\\W = nRT*ln(\frac{v_f}{v_i} )\\\\n = \frac{W}{RT*ln(\frac{v_f}{v_i} )} \\\\n = \frac{340}{8.31*300*ln(\frac{3.5 }{0.4} )}\\\\n = 0.0629 \ mol[/tex]
Therefore, the number of moles of gas present is 0.0629 mol.
