Question

Calculate the constant external pressure required to compress 2 moles of an ideal gas
from volume of 25 dm3 to 13dm3 When the work obtained is 4862.4 J.
​

Answer 1

Answer:

Calculate the work done during compression of 2 mol of an ideal gas from a volume of 1 m^3 to 10 dm^3 at 300 K against a pressure of 100 KPa.

The expression for work done is W = - 2.303 nRT log V2V1 W = - 2.303 × 2 mol × 8.314 J/mol/K × 300 × log 10 dm^31 m^3 × 1000 dm^3/m^3 W = + 22977

Answer 2

Step by step explanation:

Given : An ideal gas has 2 moles compresses from volume 25 $dm^3$ to 13

To find : The constant external pressure ($P_{ex}$).

Formula used :

$W = -P_{ex}*\Delta V$

W is the work done of gas, $P_{ex}$ is the external pressure and $\Delta V$ is the difference between final volume and initial volume.  

$\Delta V = V_2-V_1$

$V_1$ is the initial volume and $V_2$ is the final volume.

$W = - P_{ex}*(V_2-V_1)$

Conversion units used :

1. $1dm^3=1L$
2. $1J=\frac{1}{101.325} atmL$

• Calculation for $P_{ex}$

As we are given that external pressure ($P_{ex}$) of ideal gas is constant and volume is decreases so this process is Isothermal Irreversible process. Here compression is happening and we have,

n = 2 moles

$V_1=25dm^3=25L$

$V_2=13dm^3=13L$

W = 4862.4 J

Since the process is Isothermal Irreversible process so we use formula

$W = - P_{ex}*(V_2-V_1)$

to find the $P_{ex}$ rearrange above equation,

⇒  $P_{ex}=-\frac{W}{(V_2-V_1)}$

by putting values,

⇒  $P_{ex}=-\frac{4862.4J}{(13-25)L}$

           $= -\frac{4862.4J}{-12L}$

          = 405.2J/L

As we know

⇒  $1J=\frac{1}{101.325} atmL$

so the $P_{ex}$ in atm is

⇒ $P_{ex}$ =  405.2J/L

           $=405.2 * \frac{1}{101.325} atm L\frac{1}{L}$

           =3.99 atm

Hence the constant external pressure of ideal gas is 3.99 atm.