Question

Pressure versus temperature graph of an ideal gas is as shown in figure corresponding density (r) versus volume (V) graph will be –

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

Process AB is an isothermal process i.e. P µ $\frac { 1 } { \mathrm {~V} }$ and since r µ $\frac { 1 } { \mathrm {~V} }$ Ž r-V graph will be rectangular hyperbola. Pressure is increasing. Therefore volume will decrease and so density will increase. Process B-C is an isochoric process. Therefore V is constant and so r is also constant.