A perfect gas at 27°C is heated at constant pressure so as to double its volume. The increase in temperature of gas is

Question

A perfect gas at 27°C is heated at constant pressure so as to double its volume. The increase in temperature of gas is (A) 600°C (B) 327°C (C) 54°C (D) 300°C

Tardigrade Admin · Accepted Answer

We know for perfect gas V propto T Here, (V1/V2)=(T1/T2)...(i) According to question, V1=V then V2=2V and T1=300 K ∴ (1/2) =(300/T2) T2 =600 K T2 =327° C So, Δ t=327-27=300° C

Q. A perfect gas at $2 7^{\circ} C$ is heated at constant pressure so as to double its volume. The increase in temperature of gas is

$60 0^{\circ} C$

$32 7^{\circ} C$

$5 4^{\circ} C$

$30 0^{\circ} C$

We know for perfect gas
$V \propto T$
Here, $\frac{V _{1}}{V _{2}} = \frac{T _{1}}{T _{2}}$ ...(i)
According to question,
$V_{1} = V$ then $V_{2} = 2 V$ and $T_{1} = 300 K$
$∴ \frac{1}{2} = \frac{300}{T _{2}}$
$T_{2} = 600 K$
$T_{2} = 32 7^{\circ} C$
So, $Δ t = 327 - 27 = 30 0^{\circ} C$

Q. A perfect gas at 27∘C is heated at constant pressure so as to double its volume. The increase in temperature of gas is

600∘C

327∘C

54∘C

300∘C