How Do You Calculate Final Pressure in a Compressed and Cooled Ideal Gas?

AI Thread Summary
To calculate the final pressure of a compressed and cooled ideal gas, the relevant equation is P1V1/T1 = P2V2/T2. The initial values provided are P1 = 90.0 kPa, V1 = 0.0800 m3, V2 = 0.0400 m3, T1 = 300 K, and T2 = 260 K. An incorrect calculation attempt yielded P2 = 0.67 atm, but it was noted that the answer should be in kPa. The correct approach involves rearranging the equation to solve for P2, ensuring all variables are accurately applied. The final pressure calculation must adhere to the units and values specified in the problem statement.
dnl65078
Messages
14
Reaction score
0

Homework Statement



A sample of an ideal gas is both compressed and cooled. The given variables for the gas are: P1 = 90.0 kPa, V1 = 0.0800 m3, V2 = 0.0400 m3, T1 = 300. K, T2 = 260. K. Calculate the final pressure.


Homework Equations



PV=nRT
P1 V1 /T1 = P2V2 / T2


The Attempt at a Solution



P2 = T1 V / P1V T2
= 299*1.29 / 1.40 *315 * 1.29
P2 = 0.67 atm

***I thought i was doing this right, but webassign says it's wrong... please advise?
 
Physics news on Phys.org
dnl65078 said:
P1 V1 /T1 = P2V2 / T2


The Attempt at a Solution



P2 = T1 V / P1V T2

this is wrong already. mulitply your first equation by T2 and then divide by V2.
299*1.29 / 1.40 *315 * 1.29
P2 = 0.67 atm
What are all those numbers that aren't in the problem statement?
I suppose the answer should be in kPa as well.
 
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