A glass bulb is connected to an open limb manometre- The level of mercury in both limbs of the manometer was same- The bulb was heated to 57- C- If the room temperature and the atmospheric pressure were 27- C and 750 mm-Find the difference of levels in the two limbs now-A- 5-0 cmB- 7-5 cmC- 10-0 cmD- 2-5 cm

The correct option is B 7.5 cm P_1/T_1 = P_2/T_2⇒75 × d × g/300⇒(75 + x cm)dg /330 ⇒ 75 + x cm = 75 × 330/300 = 82.5 ⇒ x = 7.5 cm ...

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Byju's Answer

Standard X

Chemistry

Volume of Gases and Number of Moles

A glass bulb ...

Question

A glass bulb is connected to an open limb manometre. The level of mercury in both limbs of the manometer was same. The bulb was heated to $57^{\circ} C$ . If the room temperature and the atmospheric pressure were $27^{\circ} C a n d 750 m m$ .Find the difference of levels in the two limbs now.

2.5 cm

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5.0 cm

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7.5 cm

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10.0 cm

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Solution

The correct option is C
7.5 cm

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow \frac{75 \times d \times g}{300} \Rightarrow \frac{(75 + x c m) d g}{330} \Rightarrow 75 + x c m = \frac{75 \times 330}{300} = 82.5 \Rightarrow x = 7.5 c m$

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A glass bulb is connected to an open limb manometre. The level of mercury in both limbs of the manometer was same. The bulb was heated to 57∘C. If the room temperature and the atmospheric pressure were 27∘C and 750 mm.Find the difference of levels in the two limbs now.

The correct option is C 7.5 cm P1T1=P2T2⇒75×d×g300⇒(75+x cm)dg330⇒75 + x cm=75×330300=82.5⇒x=7.5 cm

A glass bulb is connected to an open limb manometre. The level of mercury in both limbs of the manometer was same. The bulb was heated to $57^{\circ} C$ . If the room temperature and the atmospheric pressure were $27^{\circ} C a n d 750 m m$ .Find the difference of levels in the two limbs now.

The correct option is C
7.5 cm

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow \frac{75 \times d \times g}{300} \Rightarrow \frac{(75 + x c m) d g}{330} \Rightarrow 75 + x c m = \frac{75 \times 330}{300} = 82.5 \Rightarrow x = 7.5 c m$