When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas

Question

When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas (A) will be half of original volume (B) will be 4 times the original volume (C) will be 2 times the original volume (D) will be 1 / 4 th times the original volume

Tardigrade Admin · Accepted Answer

From the combined gas equation, (p1 V1/T1)=(p2 V2/T2) (p1 V1/T1)=((1/2) p1 V2/2 T1) V1=(1/4) V2 V2=4 V1

Q. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas

will be half of original volume

will be 4 times the original volume

will be 2 times the original volume

will be 1 / 4 th times the original volume

From the combined gas equation,
$\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$
$\frac{p _{1} V _{1}}{T _{1}} = \frac{\frac{1}{2} p _{1} V _{2}}{2 T _{1}}$
$V_{1} = \frac{1}{4} V_{2}$
$V_{2} = 4 V_{1}$

Q. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas

will be half of original volume

will be 4 times the original volume

will be 2 times the original volume

will be 1 / 4 th times the original volume

From the combined gas equation, T1​p1​V1​​=T2​p2​V2​​ T1​p1​V1​​=2T1​21​p1​V2​​ V1​=41​V2​ V2​=4V1​

From the combined gas equation,
$\frac{p _{1} V _{1}}{T _{1}} = \frac{p _{2} V _{2}}{T _{2}}$
$\frac{p _{1} V _{1}}{T _{1}} = \frac{\frac{1}{2} p _{1} V _{2}}{2 T _{1}}$
$V_{1} = \frac{1}{4} V_{2}$
$V_{2} = 4 V_{1}$