If the pressure of a gas is reduced to half & its absolute temperature is doubled, then the volume of the gas will

ANS:B - increase four times.

To solve this problem, we can use the combined gas law, which states: 𝑃1⋅𝑉1𝑇1=𝑃2⋅𝑉2𝑇2T1​P1​⋅V1​​=T2​P2​⋅V2​​ Where:

• 𝑃1P1​ and 𝑃2P2​ are the initial and final pressures of the gas.
• 𝑉1V1​ and 𝑉2V2​ are the initial and final volumes of the gas.
• 𝑇1T1​ and 𝑇2T2​ are the initial and final temperatures of the gas.

Given:

• The pressure of the gas is reduced to half, so 𝑃2=12𝑃1P2​=21​P1​.
• The absolute temperature of the gas is doubled, so 𝑇2=2𝑇1T2​=2T1​.

We can substitute these values into the combined gas law: 𝑃1⋅𝑉1𝑇1=12𝑃1⋅𝑉22𝑇1T1​P1​⋅V1​​=2T1​21​P1​⋅V2​​ Now, let's solve for 𝑉2V2​: 𝑃1⋅𝑉1𝑇1=12𝑃1⋅𝑉22𝑇1T1​P1​⋅V1​​=2T1​21​P1​⋅V2​​ 𝑉2=2𝑃1⋅𝑉112𝑃1⋅2𝑇1V2​=21​P1​⋅2T1​2P1​⋅V1​​ 𝑉2=2𝑉112V2​=21​2V1​​ 𝑉2=2×2×𝑉1V2​=2×2×V1​ 𝑉2=4𝑉1V2​=4V1​ So, the volume of the gas will increase four times. Therefore, the correct answer is: increase four times.