Certain amount of a gas occupies a volume of 0-4 litre at 17- C- To what temperature should it be heated so that its volume gets reduced to half-

V_1T_1= V_2T_2 Here, V_1= 0.4L, T_1= 273+17= 290K, V_2= 0.42= 0.2L Then, T_2= 0.2 × 2900.4 =145K = 145 273 = 128^oC ...

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Standard IX

Chemistry

Temperature-Volume Relationship

Certain amoun...

Question

Certain amount of a gas occupies a volume of 0.4 litre at $17^{\circ} C$ . To what temperature should it be heated so that its volume gets reduced to half?

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Solution

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{1} = 0.4 L, T_{1} = 273 + 17 = 290 K, V_{2} = \frac{0.4}{2} = 0.2 L$
Then, $T_{2} = \frac{0.2 \times 290}{0.4} = 145 K = 145 - 273 = - 128^{o} C$

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Certain amount of a gas occupies a volume of 0.4 litre at 17∘C. To what temperature should it be heated so that its volume gets reduced to half?

V1T1=V2T2Here, V1=0.4L,T1=273+17=290K,V2=0.42=0.2LThen, T2=0.2×2900.4=145K=145−273=−128oC

Certain amount of a gas occupies a volume of 0.4 litre at $17^{\circ} C$ . To what temperature should it be heated so that its volume gets reduced to half?

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Here, $V_{1} = 0.4 L, T_{1} = 273 + 17 = 290 K, V_{2} = \frac{0.4}{2} = 0.2 L$
Then, $T_{2} = \frac{0.2 \times 290}{0.4} = 145 K = 145 - 273 = - 128^{o} C$