A gas at temperature 250 K is contained in a closed vessel- If the gas is heated and the temperature rises by 1- C- the percentage increase in its pressure is nearlyA- 0-4 -B- 0-6 -C- 0-8 -D- 1-0 -

The correct option is A 0.4 %We know, nbsp; if mass and volume (V) are constant, nbsp; P ∝ T ⇒PT = constant ⇒P_1T_1 = P_2T_2 ⇒P_2P_1 = T_2T_1 ⇒P_2 P_1P_1 ...

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Standard XII

Physics

P-T Relationship

A gas at temp...

Question

A gas at temperature $250 K$ is contained in a closed vessel. If the gas is heated and the temperature rises by $1^{\circ} C,$ the percentage increase in its pressure is nearly

0.4 %

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0.6 %

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0.8 %

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1.0 %

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Solution

The correct option is A $0.4 %$
We know,
if mass and volume $(V)$ are constant,
$P \propto T$
$\Rightarrow \frac{P}{T} =$ constant
$\Rightarrow \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\Rightarrow \frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{P_{2} - P_{1}}{P_{1}} = \frac{T_{2} - T_{1}}{T_{1}}$
$\Rightarrow \frac{P_{2} - P_{1}}{P_{1}} = \frac{1}{250}$
$\Rightarrow \frac{P_{2} - P_{1}}{P_{1}} % = \frac{100}{250} = \frac{2}{5} = 0.4 %$

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A gas at temperature 250 K is contained in a closed vessel. If the gas is heated and the temperature rises by 1∘C, the percentage increase in its pressure is nearly

The correct option is A 0.4%We know, if mass and volume (V) are constant, P∝T ⇒PT= constant ⇒P1T1=P2T2 ⇒P2P1=T2T1 ⇒P2−P1P1=T2−T1T1 ⇒P2−P1P1=1250 ⇒P2−P1P1%=100250=25=0.4%

A gas at temperature $250 K$ is contained in a closed vessel. If the gas is heated and the temperature rises by $1^{\circ} C,$ the percentage increase in its pressure is nearly