Question

In a process the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas increases, then work done by the gas:

• A. Is positive
• B. Is negative
• C. Is zero
  
• D. may be positive
  

Answer 1

The correct option is B

Is negative

Given,
$P\propto \frac{1}{V^2}$
$\Rightarrow P=\frac{k}{V^2}V\Rightarrow P{V}^2+k$

$\Rightarrow PV\cdot V=k\Rightarrow nRTV=k\Rightarrow TV=k_1$
Since temperature increases therefore volume decreases.
Hence, the work done by the gas should be negative.
Final answer: (b)