What is the temperature increase of the gas?

AI Thread Summary
The discussion focuses on calculating the temperature increase of gas in a rubber raft after work is done on it by an air pump. The work done on the gas is determined to be 0.32 Joules, calculated using the formula Work = Force * Distance. For the temperature increase, the user attempts to apply the ideal gas law and thermal energy equations, using specific values for air density and specific heat capacity. The calculations involve determining mass from volume and density, and then using the work done to find the change in temperature. Ultimately, the user seeks clarification on the correct approach to derive the temperature increase from the work done.
Tori Grafe
Messages
1
Reaction score
0

Homework Statement


In an empty rubber raft the pressure is approximately constant. You push on a large air pump that pushes 1.0L (1.0×10^−3 m^3) of air into the raft. You exert a 16N force while pushing the pump handle 2.0×10^−2 m .

Part A: Determine the work done on the gas.
Part B: If all of the work is converted to thermal energy of the 1.0 L of gas, what is the temperature increase of the gas? Assume that the air obeys the ideal gas law and is initially at 293 K.

2. Homework Equations

Just trying to come up with some equations that may help, although I have had no luck...
P*V=n*R*T
Work = -n*R*deltaT
W= (-/+) deltaP*V
Delta T = Q/c*m
Uthermal = 3/2*n*R*T
delta U = c*m*delta T

The Attempt at a Solution


Part A: I have already solved this part as you can see below...
Work = Force * Distance
Work = 16 N * 2.0x10^-2 m
Work = .32 N*m
Work = .32 (kg*m^2)/s^2
Work = .32 J

Part B: I have attempted a couple of times as you can see below... I really need help on this
Work = Q
Delta T = Q/c*m
m = ro * v
ro = 1.3 kg/m^3
v= 1.0x10^-3 m^3
c= 700 J/(kg * degrees C)
m = 1.3x10^-3 kg

delta T = .32J / (700 (J/(kg * degrees C)) * 1.3x10^-3 kg)

Work = -delta P * v
.32 J = -delta P * 1.0x10^-3
-delta P = 320 N/m^2
320 N/m^2 = n*R*delta T
320 N/m^2 = n*(8.314 J/mol * K)*delta T
 
Physics news on Phys.org
I think the equation
ΔU=nCvΔT
Might help you.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top