Question

What is the amount of work done when two moles of ideal gas are compressed from a volume of \[1{m^3}\] to \[10d{m^3}\] at \[300K\] against a pressure of \[100kPa\]?
A. \[99kJ\]
B. \[ - 99kJ\]
C. \[114.9kJ\]
D. \[ - 114.9kJ\]

Answer 1

Hint: The work done by ideal gas can be calculated from the pressure and volume of an ideal gas. The change in volume must be considered. When a gas is compressed, there will be a change in volume. Thus, pressure and change in volume must be considered while calculating the work done.

Formula used:
\[W = P\Delta V\]
\[W\] is work done
\[P\] is pressure in pascals
\[\Delta V\] is change in volume

Complete step by step solution:
Given that Two moles of ideal gas are compressed from a volume of \[1{m^3}\] to \[10d{m^3}\] at \[300K\] against a pressure of \[100kPa\].
The final volume \[{V_2}\] is \[10d{m^3}\]the unit is in \[d{m^3}\] it should be converted to \[{m^3}\].
\[10d{m^3} = \dfrac{{10}}{{1000}}{m^3} = \dfrac{1}{{100}}{m^3}\]
Thus, the final volume \[{V_2}\] is \[\dfrac{1}{{100}}{m^3}\]
Initial volume \[{V_1} = 1{m^3}\]
Given pressure is \[100kPa\], the units of pressure were in kilopascals, it should be converted into pascals.
\[1kPa = 1000Pa\]
Thus, the pressure is \[1000 \times 100Pa = {10^5}Pa\]
Substitute the values of pressure and volume in the formula of work done
\[W = {10^5} \times \left( {\dfrac{1}{{100}} - 1} \right)\]
On simplification, the work done will be
\[W = - 99000J = - 99kJ\]
The amount of work done when two moles of ideal gas are compressed from a volume of \[1{m^3}\] to \[10d{m^3}\] at \[300K\] against a pressure of \[100kPa\] is \[99kJ\].
Option A is correct.

Note:
When an ideal gas is compressed by an external pressure and the change in volume is less than zero. The work done will be positive. In the above solution the change in volume is negative. Thus, the work done will be positive. The work done will be \[99kJ\] but not \[ - 99kJ\].