Question

2 litre of N2 at 0 degree Celsius and 5 ATM are expanded isothermally against a constant external pressure of 1 ATM until the pressure of gas reaches 1 ATM assuming the gas to be ideal calculate work of expansion?

Answer 1

Answer:- work of expansion is 506.6 J.

Solution:- Initially the gas is at 5 atm pressure and it's volume is 2 liter. The final pressure is going to be decreased to 1 atm.  Process is isothermal means the temperature is constant. Using Boyle's law we calculate the final volume of the gas:

$P_1V_1=P_2V_2$

Given, $P_1$ = 5 atm

$P_2$ = 1 atm

$V_1$ = 2 L

$V_2$ = ?

Let's plug in the values in the equation to calculate final volume:

$5atm(2L)=1atm(V_2)$

$V_2=\frac{5atm*2L}{1atm}$

$V_2$ = 10 L

External pressure is also given as 1 atm. Pressure-volume work is calculated using the formula:

$w=-P_e_x_t\Delta V$

Let's plug in the values in it:

$w=-1atm(10L-5L)$

$w=-5atm.L$

since, $1atm.L=101.325J$

So, $-5atm.L(\frac{101.325J}{1atm.L})$

= -506.6 J

Negative sign indicates the work is done by the system. So, we could say that work of expansion is 506.6 J.

Answer 2

W = – Pext(V2  –  V1)

= – 1 × (V2 – V1)

As given, P1 = 5 atm, P2 = 1 atm, V1 = 2 litre and V2 = ?

Using, P1V1 = P2V2

5 × 2 = 1 × V2

V2 = 10 litre

Therefore, W = – 1 × (10 – 2)

=– 8 litre atm

= – (8×1.987)/0.0821 calorie

= – (8×1.987×4.184)/0.0821 J

Hence, work of expansion = 810.09 J