Question

Volume of a gas is expanded from 3 dm cube to 6dm cube against a constant pressure of 1 atmosphere. calculate the work done by the system

Answer 1

Since it is expansion, the work will be negative in case of Chemistry.

Now, work at constant pressure = pressure × change in volume

Change in volume= final volume - initial volume = (6-3)dm3 = 3 dm3

Work= 1atm × 3dm cube= 3 atm-dm cube = 3 L atm

I hope it's correct.

Answer 2

Answer:

The work done by the system calculated is $-303.99J$.

Explanation:

Given data,

The initial volume of gas before expansion, $V_{i}$ = $3dm^{3}$ = $3L$

The final volume of gas after expansion, $V_{f}$ = $6dm^{3}$ = $6L$

The constant pressure of the gas, P = $1atm$

The work done by the system =?

As given,

• The expansion of volume is occurring, i.e., work is done by the system.

And,

• Work done by the system is always negative.

Now, work done is given by:

• W = $P\Delta V$

As discussed above, work done = negative.

Therefore,

• W = $-P\Delta V$
• W = $-P ( V_{f} -V_{i} )$
• W = $-1( 6-3)$
• W = $-3atm-L$

Now, we have to convert atm-L into J.

• $1atm-L = 101.33J$
• $-3atm-L = (-3\times 101.33)J$ = $-303.99J$

Hence, the work done by the system calculated is $-303.99J$.