If the absolute temperature of an ideal gas is tripled and simultaneously the pressure is reduced to one third; then the volume of gas will

ANS:B - increase nine fold

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 𝑇1T1​ are the initial pressure and temperature,
• 𝑉1V1​ is the initial volume,
• 𝑃2P2​ and 𝑇2T2​ are the final pressure and temperature,
• 𝑉2V2​ is the final volume.

Given:

• Initial pressure (𝑃1P1​) is reduced to one third (1331​).
• Initial temperature (𝑇1T1​) is tripled (3𝑇3T).

Let's represent the initial state as subscript "1" and the final state as subscript "2". So, we have:

• 𝑃1=𝑃P1​=P
• 𝑇1=𝑇T1​=T
• 𝑃2=𝑃3P2​=3P​
• 𝑇2=3𝑇T2​=3T

Now, we can rearrange the combined gas law equation to solve for 𝑉2V2​: 𝑉2=𝑉1⋅𝑃1⋅𝑇2𝑃2⋅𝑇1V2​=V1​⋅P2​⋅T1​P1​⋅T2​​ 𝑉2=𝑉⋅𝑃⋅(3𝑇)𝑃3⋅𝑇V2​=V⋅3P​⋅TP⋅(3T)​ 𝑉2=𝑉⋅3𝑃⋅𝑇𝑃3⋅𝑇V2​=V⋅3P​⋅T3P⋅T​ 𝑉2=𝑉⋅9𝑃⋅𝑇𝑃⋅𝑇V2​=V⋅P⋅T9P⋅T​ 𝑉2=𝑉⋅9V2​=V⋅9 So, the volume of the gas will increase nine-fold. Therefore, the correct answer is "increase nine fold".