Another physics problem that I trip over...

Test Preparation AP Sciences
1

<p>I am very confuse over these question. It will be great if you can help. Never taken physics before
Here are the equation involved:
PV=nRT
Q=mcΔT
U=3/2k(b)T
Vrms=square root √3RT/√M
W=-PΔV
ΔU=Q+W</p>

<p>Here are the questions:

  1. Assuming n and R are both held constant, what happens to T if P is doubled and V is tripled?
  2. Calculate m if c= 4000 J/kgC, Q= 6.2 KJ, and T=12C. To do this correctly KJ needs to be converted into the units of J.
  3. If U doubles and kb, R, and M remain the same values, how does Vrms change?
  4. If ΔV is positive and ΔU is zero, what is the sign of Q? Justify your answer using the last two equations.</p>

<p>Thanks</p>

2

<ol>
<li><p>Using PV=nRT and the assumptions that both n and R are held constant, we know that the equation changes to (2P)(3V)=nRT, which is the same as 6PV=nRT. We know that the equation represents equilbrium, so both sides of the equation must equal each other. </p></li>
<li><p>Since we have the equation Q = mcΔT, we can substitute values in for the equation: 6.2 kJ = m(4000 J/kgC)(12C). Now, 1 J is equal to 0.001 kJ so we can change the equation to 6200 J = m(4000 J/kgC)(12C).</p></li>
</ol>

<p>*Note that the units for ΔT, which can be Celcius or Kelvin (K = C + 273) depends on c and can vary between problems. </p>

<ol>
<li><p>Using both U = 3/2k(b)T and Vrms = (sqrt)3RT/m, we know from the information that U doubles, making the first equation 2U = 3/2k(b)T. T doubles due to U doubling and k(b) staying constant. Since we now know that T is doubled, Vrms = (sqrt)3R(2T)/m.</p></li>
<li><p>Since we know ΔV is positive, we can input it into the equation W =-PΔV. If ΔV is positive, W MUST be negative (due to multiplying negative P and positive V). Now we know we know ΔU = 0 and W is negative in ΔU = Q + W, making it 0 = Q + (-W).</p></li>
</ol>

<p>All the information should be enough to get you the answer. The reason I didn’t give you the answer is because you need to understand Physics as it is NOT a simple plug-in subject. If you have any more questions, let me know.</p>

3

<p>These are extremely basic… most of these are plug in-find answer questions.
I suggest reading the textbook.</p>