Calculating Gas Temperature Change Using the Ideal Gas Law

AI Thread Summary
The discussion revolves around calculating the new temperature of a gas after it is compressed using the ideal gas law. The initial conditions are a temperature of 293.15 K and atmospheric pressure, with the gas compressed to a volume one-fifteenth of its original size and a pressure of 3000 kPa. The user initially misapplies the formula, leading to confusion about the correct temperature calculation. After clarification, they realize the correct rearrangement of the equation yields a final temperature of 307 K. The conversation highlights the importance of careful manipulation of equations in thermodynamics.
turnip
Messages
30
Reaction score
0

Homework Statement


A gas at 293.15 degrees kelvin and atmospheric pressure is compressed to a volume one fifteenth as large as its original volume and absolute pressure of 3000kPa. What is the new temperature of the gas?


Homework Equations


p1v1/t1=p2v2/t2


The Attempt at a Solution


if i rearrange the equation above for t2 i get t2=101/293.15 x 3000 x 1/15= 1.722x10-3

if i do it the wrong way and flip the divisions around so that t2=293.15 x 3000 x 1/15/101 =580.49

i take the 580 - 273 =307

307 is the right answer. perhaps i am doing it the right way, i just don't understand how this could be correct
please explain this to me
 
Physics news on Phys.org
turnip said:
if i rearrange the equation above for t2 i get t2=101/293.15 x 3000 x 1/15= 1.722x10-3

u r getting it 1/t2 and not t2.. verify urself.
 
haha can't believe i didnt get that
i need more sleep :P
 
oh and thanks :)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top