Question

The pressure of an ideal gas undergoing a quasi static process is inversely proportional to the square of 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

According to question,

$p\propto \frac{1}{V^2}$

$\Rightarrow p=\frac{k}{V^2}$ (where k is constant)

$\Rightarrow p{V}^2=k$

$\Rightarrow \left(pV\right)\left(V\right)=k$

Using the equation of ideal gas

$\Rightarrow \left(nRT\right)\left(V\right)=k$

$\Rightarrow T\propto \frac{1}{V}(n,R\&k\text{are constant)}$

From above expression, temperature of gas increases, volume of gas will decrease

i.e. $V_f<V_i$

Hence, for the compression of gas, work done is negative

Why this question? Tip : In this problem, in order to compare $W$ we must obtain relation for $T$ vs $V$. This can be achieved by combining equation of state with given process equation