A gas at 0°C was subjected to constant pressure cooling until its volume became half the original volume. The temperature of the gas at this stage will be

ANS:C - -136.5°C

To solve this problem, we can use Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature measured on the Kelvin scale. Mathematically, it can be expressed as: 𝑉1𝑇1=𝑉2𝑇2T1​V1​​=T2​V2​​ where:

• 𝑉1V1​ and 𝑇1T1​ are the initial volume and temperature of the gas,
• 𝑉2V2​ and 𝑇2T2​ are the final volume and temperature of the gas.

Given that the gas is initially at 0°C (which is 273.15 K) and its volume becomes half the original volume, we can set up the equation as follows: 𝑉1𝑇1=𝑉2𝑇2T1​V1​​=T2​V2​​ 1273.15=1/2𝑇2273.151​=T2​1/2​ Solving for 𝑇2T2​: 𝑇2=(1/2)×273.15T2​=(1/2)×273.15 𝑇2=136.575 𝐾T2​=136.575K So, the temperature of the gas at this stage will be 136.575 𝐾136.575K. Converting this temperature to Celsius: 𝑇2°𝐶=𝑇2𝐾−273.15T2°C​=T2K​−273.15 𝑇2°𝐶=136.575−273.15T2°C​=136.575−273.15 𝑇2°𝐶=−136.575 °𝐶T2°C​=−136.575°C Therefore, the correct answer is: (c) -136.5°C