Question

A sample of nitrogen gas expands in volume from 1.6 to 5.4 l at constant temperature. calculate the work done in joules if the gas expands

Answer 1

The question is incomplete, here is the complete question:

A sample of nitrogen gas expands in volume from 1.6 to 5.4 L at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of 0.80 atm, and (c) against a constant pressure of 3.7 atm. ( 1 L.atm = 101.3 J)

Answer:

For a: The work done for the given process is 0 J

For b: The work done for the given process is -308.04 J

For c: The work done for the given process is -1424.7 J

Step-by-step explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

`W=-P\Delta V=-P(V_2-V_1)` ......(1)

W = amount of work done

P = pressure

`V_1` = initial volume

`V_2` = final volume

To convert this into joules, we use the conversion factor:

`1L.atm=101.33J`

• For a:

At vacuum, the pressure of the system will be 1 atm

We are given:

`P=0atm\\V_1=1.6L\\V_2=5.4L`

Putting values in above equation, we get:

`W=-0atm* (5.4-1.6)L=0L.atm=0J`

Hence, the work done for the given process is 0 J

• For b:

We are given:

`P=0.8atm\\V_1=1.6L\\V_2=5.4L`

Putting values in above equation, we get:

`W=-0.8atm* (5.4-1.6)L=-3.04L.atm=-308.04J`

Hence, the work done for the given process is -308.04 J

• For c:

We are given:

`P=3.7atm\\V_1=1.6L\\V_2=5.4L`

Putting values in above equation, we get:

`W=-3.7atm* (5.4-1.6)L=-14.06L.atm=-1424.7J`

Hence, the work done for the given process is -1424.7 J

Answer 2

Missing part in the question:
if the gas expands a) against a vacuum b) against a constant pressure of 0.80 atm and c) against a constant pressure of 3.7 atm

Solution:
For a gas transformation at constant temperature, if the pressure p is constant, the work done is equal to

`W=p \Delta V`
where
`\Delta V` is the change in volume of the gas.
In our problem,

`\Delta V=5.4 L-1.6 L=3.8 L = 3.8 \cdot 10^(-3)m^3`

a) When the gas expands in vacuum, the pressure is zero: p=0 atm, so the work done is zero:

`W=p \Delta V=( 0)(3.8 \cdot 10^(-3)m^3)=0`

b) The pressure is
`p=0.80 atm= 8.1 \cdot 10^(4)Pa`, so the work done is

`W=p \Delta V=( 8.1 \cdot 10^(4)Pa)(3.8 \cdot 10^(-3)m^3)=307.8 J`

c) The pressure is
`p=3.7 atm= 3.7 \cdot 10^(5)Pa`, so the work done is

`W=p \Delta V=( 3.7 \cdot 10^(5)Pa)(3.8 \cdot 10^(-3)m^3)= 1420.1 J`