Question:

The volume of a gas expands by $0.25 \,m ^{3}$ at a constant pressure of $10^{3} \,N / m$. The work done is equal to

Updated On: Aug 1, 2022
  • 250 watt
  • 2.5 erg
  • 250 newton
  • 250 joule
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The Correct Option is D

Solution and Explanation

Given: Change in volume of a gas $(dV) =0.25\, m ^{3}$ and constant pressure $(P)=10^{3}\, N / m ^{2}$. We know that work done $(W)=P \cdot d V=10^{3} \times 0.25$ $=250$ joule.
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Concepts Used:

Thermal Expansion

Thermal expansion is the tendency of matter to change its shape, area, and volume in response to a change in temperature. Temperature is a monotonic function of the average molecular kinetic energy of a substance.

The expansion of the solid material is taken to be the linear expansion coefficient, as the expansion takes place in terms of height, thickness and length. The gaseous and liquid expansion takes the volume expansion coefficient. Normally, if the material is fluid, we can explain the changes in terms of volume change. 

The bonding force among the molecules and atoms differs from material to material. These characteristics of the compounds and elements are known as the expansion coefficient.

thermal expansion