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

`n=0.033mole`

Step-by-step 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(V_2)/(V_1)`

`n=(W)/(RTIn(V_2)/(V_1))`

`n=(210)/(8.32*300In(5.0)/(0.4))`

`n=0.033mole`