Question

In Boyles experiment for a given gas at different temperatures the graph drawn between pressure and density are straight lines as shown then:

• A. $T_1>T_2$
• B. $T_2>T_1$
• C. $T_1=T_2$
• D. $T_1^3=T_2$

Answer 1

The correct option is A $T_1>T_2$

First we derive an expression for density of a gas.
Density $=$ $\frac{mass\left(w\right)}{volume\left(V\right)}$
but we know that PV$=$nRT
now, n $=$ $\frac{w}{M}$ where M is molecular weight
So, PV $=$ $\frac{w}{M}RT$
this gives $\frac{w}{V}=\frac{PM}{RT}$ $=$ density $=$ d
so, $P=\frac{dRT}{M}$
so pressure vs density graph will be a straight line with slope $=$ $\frac{RT}{M}$
Hence, steeper the line, more is slope and so more is temperature.
So here $T_1>T_2$