Question

Work is required to compress 5.00 mol of air at 20.00C and 1.00 atm to one-tenth of the original volume by an adiabatic process. How much work is required to produce this same compression?

Answer 1

Answer:46.03 KJ

Step-by-step explanation:

no of moles(n)=5

temperature of air(T)=
`20^(\circ)`

pressure(p)=1 atm

final volume is
`(V)/(10)`

We know work done in adaibatic process is given by

W=
`(P_iV_i-P_fV_f)/(\gamma -1)`

`\gamma` for air is 1.4

we know
`P_iV_i^(\gamma )= P_fV_f^(\gamma )`

`1\left ( V\right )^(\gamma )`=
`P_f\left ((v)/(10)\right )^(\gamma )`

`10^(\gamma )=P_f`

`P_f=25.118 atm`

W=
`(1* 0.1218-25.118* 0.01218)/(1.4 -1)`

W=-46.03 KJ

it means work is done on the system