A gas of volume 100 cc is kept in a vessel at pressure 10-4 Pa maintained at temperature 24- C- Now- if the pressure is increased to 105 Pa- keeping the temperature constant- then the volume of the gas becomes-

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Standard XII

Physics

Thermodynamic Processes

A gas of volu...

Question

A gas of volume 100 cc is kept in a vessel at pressure 10.4 Pa maintained at temperature $24^{\circ} C$ . Now, if the pressure is increased to 105 Pa, keeping the temperature constant, then the volume of the gas becomes:

10 cc

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100 cc

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1 cc

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1000 cc

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Solution

The correct option is A 10 cc
Given

$V_{1} = 100 c c .$

$P_{1} = 10.4 P a$

$V_{2} = ?$

$P_{2} = 105 P a$

$P_{1} V_{1} = P_{2} V_{2}$

$∴ V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{105 \times 100}{10.4} = 10 c c$

Hence, the correct option is $A$

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A gas of volume 100 cc is kept in a vessel at pressure 10.4 Pa maintained at temperature 24∘C. Now, if the pressure is increased to 105 Pa, keeping the temperature constant, then the volume of the gas becomes:

The correct option is A 10 ccGivenV1=100 cc.P1=10.4PaV2=?P2=105PaP1V1=P2V2∴V2=P1V1P2=105×10010.4=10ccHence, the correct option is A

A gas of volume 100 cc is kept in a vessel at pressure 10.4 Pa maintained at temperature $24^{\circ} C$ . Now, if the pressure is increased to 105 Pa, keeping the temperature constant, then the volume of the gas becomes:

The correct option is A 10 cc
Given

$V_{1} = 100 c c .$

$P_{1} = 10.4 P a$

$V_{2} = ?$

$P_{2} = 105 P a$

$P_{1} V_{1} = P_{2} V_{2}$

$∴ V_{2} = \frac{P_{1} V_{1}}{P_{2}} = \frac{105 \times 100}{10.4} = 10 c c$

Hence, the correct option is $A$