Question

For many purposes we can treat propane (CzH) as an ideal gas at temperatures above its boiling point of -42°C. Suppose the temperature of a sample of propane gas is raised from -20.0 °C to 18.0 °C, and at the same time the pressure is decreased by 10.0%. increase Does the volume of the sample increase, decrease, or stay the same? decrease x 5 ? stays the same If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent.

Answer 1

sample of propane gas is Ans: - Given that The temperature of raised from -20°c out the same time a to 18°c the pressure is decreased by 10% for ideal gas we know the equation PIV we know that = P2 V 2 T 1. Tz -accooding to the given dat a T, = -20C = 273-206 = 253K Ty : 18c 2 73+ if : 291K . pressure is decreased by 104. So 904. - 0.9071 .: Pa- 0.9021 These verlues are replace the above equation divi - 0.90P V2 253K 29114 291 K -1 .277 253kx 0.90 so here we obtained so Voz is grater 2 = 11277 them vi [ V2 sv.]

The roume of the sample increases we have - 1.277 12 = 1. 27V V2 -VI = 1.211V, -V, --[4-277-1] V2-V = m [0217] v2 = 0.211 8 percent increase in volume Chorge in volume V2v noo = 0.217 X100 | ۷۱ initial - 27.7% .. percentage change in volume = 27.70).