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

Explanation:

no of moles(n)=5

temperature of air(T)=$20^{\circ}$

pressure(p)=1 atm

final volume is $\frac{V}{10}$

We know work done in adaibatic process is given by

W=$\frac{P_iV_i-P_fV_f}{\gamma -1}$

$\gamma$ for air is 1.4

we know $P_iV_i^{\gamma }= P_fV_f^{\gamma }$

$1\left ( V\right )^{\gamma }$=$P_f\left (\frac{v}{10}\right )^{\gamma }$

$10^{\gamma }=P_f$

$P_f=25.118 atm$

W=$\frac{1\times 0.1218-25.118\times 0.01218}{1.4 -1}$

W=-46.03 KJ

it means work is done on the system