Question

It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present

Answer 1

Answer:

$n=0.033mole$

Explanation:

From the question we are told that:

Initial volume $V_1=0.40L$

Final Volume$V_2=5.0L$

Work $W=210J$

Temperature $T=300k$

Generally the equation for Ideal gas is mathematically given by

$W=nRTIn\frac{V_2}{V_1}$

$n=\frac{W}{RTIn\frac{V_2}{V_1}}$

$n=\frac{210}{8.32*300In\frac{5.0}{0.4}}$

$n=0.033mole$