Calculating the Value of Q for Constant Pressure Heat Transfer

AI Thread Summary
The value of Q in the problem is given as 29.1 kJ, which equals 29100 J. At constant pressure, the gas expands, and not all heat energy contributes to raising the internal energy, as some is used for work done during expansion. In contrast, at constant volume, all heat energy goes into increasing internal energy. The relationship between Q and internal energy (U) is clarified, showing that while Q is equal to the heat transferred, U changes due to work done. The calculations confirm that using Q leads to a sensible ratio of specific heats, C_p to C_v.
annalise17
Messages
10
Reaction score
0
1. What is the value of Q in this problem? Is it a value that's stated in the question or does it need calculating?

A sample consisting of 5 moles of an ideal gas at a temperature of 300 K and a
pressure of 1.00 × 105 Pa is heated to a temperature of 500 K at constant pressure.
The amount of heat transferred to the gas is 29.1 kJ.

Homework Equations


No equations needed really, I just need to make sure I'm using the right values!


The Attempt at a Solution


Is Q = 29.1 kJ = 29100 J (or would that be U?)
Is Q the same value for constant volume and constant pressure while ΔU changes because ΔU = Q + W and there is no work in a constant volume situation?
 
Physics news on Phys.org
In this example the gas is at constant pressure which means that it will expand and therefore external work is done during the expansion.
All of the heat energy does not go to raise the internal energy (raise the temperature) of the gas
If the gas is heated at constant volume no external work is done ands all of the heat energy goes to internal energy and raise the temperature.
 
I got that bit, I just wasn't sure whether we were actually given Q or U but I've worked it out:

∆U = Q = 29100 J
∆U = C_v (T_2-T_1)
etc. Using Q as that gave me a sensible ratio of C_p to C_v so I'm happy!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top