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

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

Tardigrade Admin · Accepted Answer

P propto (1/V2) ⇒ P=(k/V2) V ⇒ P V2=k ⇒ P V. V=k ⇒ n R T V=k ⇒ T V=k1 Since temperature increases therefore volume decrease. Hence the work done by the gas should be negative.

Q. 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

is positive

is negative

is zero

may be positive

$P \propto \frac{1}{V ^{2}}$
$\Rightarrow P = \frac{k}{V ^{2}} V$
$\Rightarrow P V^{2} = k$
$\Rightarrow P V . V = k$
$\Rightarrow n RT V = k$
$\Rightarrow T V = k_{1}$
Since temperature increases therefore volume decrease.
Hence the work done by the gas should be negative.

Q. 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

is positive

is negative

is zero

may be positive

P∝V21​ ⇒P=V2k​V ⇒PV2=k ⇒PV.V=k ⇒nRTV=k ⇒TV=k1​ Since temperature increases therefore volume decrease. Hence the work done by the gas should be negative.

$P \propto \frac{1}{V ^{2}}$
$\Rightarrow P = \frac{k}{V ^{2}} V$
$\Rightarrow P V^{2} = k$
$\Rightarrow P V . V = k$
$\Rightarrow n RT V = k$
$\Rightarrow T V = k_{1}$
Since temperature increases therefore volume decrease.
Hence the work done by the gas should be negative.