If pressure of a gas contained in a closed vessel is increased by 0.4 % when heated by 1° C , its initial temperature must be

Question

If pressure of a gas contained in a closed vessel is increased by 0.4 % when heated by 1° C , its initial temperature must be (A) 250 K (B)  250° C  (C) 2500 K (D)  25° C

Tardigrade Admin · Accepted Answer

(pV/T)= constant; V = constant ∴ (p/T)= constant ∴ (p/T)=(100.4 p/100)× (1/(T+1)) (1/T)=(100.4/100)× (1/(T+1)) (1/T)=(100.4/100T+100) T = 250 K

Q. If pressure of a gas contained in a closed vessel is increased by $0.4%$ when heated by $1^{\circ} C$ , its initial temperature must be

250 K

$25 0^{\circ} C$

2500 K

$2 5^{\circ} C$

$\frac{p V}{T} =$ constant; $V =$ constant
$∴ \frac{p}{T} =$ constant
$∴ \frac{p}{T} = \frac{100.4 p}{100} \times \frac{1}{( T + 1 )}$
$\frac{1}{T} = \frac{100.4}{100} \times \frac{1}{( T + 1 )}$
$\frac{1}{T} = \frac{100.4}{100 T + 100}$
$T = 250 K$

Q. If pressure of a gas contained in a closed vessel is increased by 0.4% when heated by 1∘C , its initial temperature must be

250 K

250∘C

2500 K

25∘C

TpV​= constant; V= constant ∴Tp​= constant ∴Tp​=100100.4p​×(T+1)1​ T1​=100100.4​×(T+1)1​ T1​=100T+100100.4​ T=250K

Q. If pressure of a gas contained in a closed vessel is increased by $0.4%$ when heated by $1^{\circ} C$ , its initial temperature must be

$25 0^{\circ} C$

$2 5^{\circ} C$

$\frac{p V}{T} =$ constant; $V =$ constant
$∴ \frac{p}{T} =$ constant
$∴ \frac{p}{T} = \frac{100.4 p}{100} \times \frac{1}{( T + 1 )}$
$\frac{1}{T} = \frac{100.4}{100} \times \frac{1}{( T + 1 )}$
$\frac{1}{T} = \frac{100.4}{100 T + 100}$
$T = 250 K$