Question

For a closed (non rigid) container containing $n=10\text{moles}$ of an ideal gas fitted with movable, frictionless, weightless piston operating such that pressure of gas remains constant at $0.821\text{atm}$, which graph represents correct variation of $\log V\text{vs}\log T$ where $V$ is in litre and $T$ in kelvin.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We know,
$PV=nRT$
Substituting the values, we get
$\Rightarrow 0.821\times V=10\times 0.0821\times T$
$\Rightarrow V=T$
$\Rightarrow \log V=\log T+\mathrm{0...}\left(1\right)$
We know that equation of a straight line is $y=mx+c...\left(2\right)$
Comparing equation $\left(1\right)$ and $\left(2\right)$, for a graph of $\log V\text{vs}\log T$, we get
$\text{Intercept (c)}=0$
$\text{Slope}=1$
$\Rightarrow \tan \left(\theta \right)=1$
$\Rightarrow \theta ={45}^{\circ }$