Question 1 Report
A given mass of gas at a temperature of 30°C is trapped in a tube of volume V. Calculate the temperature of the gas when the volume is reduced to two-third of its original value by applying a pressure twice the original value.
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law is given by: (P1 x V1) / T1 = (P2 x V2) / T2 where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature of the gas. In this problem, we know that the initial temperature is 30°C, or 303 K (since temperature must be in Kelvin for gas laws). We also know that the initial volume is V, and the final volume is two-thirds of the initial volume, or (2/3)V. Finally, we know that the final pressure is twice the initial pressure, or 2P. Substituting these values into the combined gas law, we get: (P x V) / 303 = (2P x 2V/3) / T2 Simplifying this equation, we get: T2 = (2P x 2V/3 x 303) / P x V Canceling out the P and V terms, we get: T2 = (2 x 2/3 x 303) / 1 Simplifying this equation, we get: T2 = 404 K Converting this temperature back to Celsius, we get: T2 = 404 - 273 = 131°C Therefore, the temperature of the gas when the volume is reduced to two-thirds of its original value by applying a pressure twice the original value is 131°C. The correct option is 131oC.