Recent content by Kakashi

At constant pressure we have: $$ dq_{p}=dH=C_{p} dT $$ which implies The enthalpy change dH reflects the increase in internal energy associated with molecular motions such as rotational and vibrational modes. The atoms, molecules or ions that compose a system can undergo several types of...

I am quite confused by the way my textbook presents the Joule–Thomson experiment. A gas flowing through a throttling valve from a high pressure Pi to a lower pressure Pf is an open system. However, the book treats the gas as if it were a closed system and imagines the upstream side acting like a...

A class of n students takes a test with m questions. Student i submits answers to the first $$ m_{i} $$ questions. Let $$M= \sum_{i=1}^{n} m_{i}=m_{1}+m_{2}+..m_{n} $$ denote the total number of submitted answers . The grader randomly picks one submitted answer. We define two discrete random...

a) If the buffer starts empty the number of packets stored can be any integer $$ 0 \leq k_{1} \leq b $$. At the end of the second slot the system transmits either $$ k_{1} $$ or c depending which is less. If $$ k_{1}\geq c $$ the number of packets remaining in the buffer is $$ k_{1}-c $$ and if...

a) The kth symbol can be 0 or 1. If 0 or 1 are transmitted they can either be received correctly or incorrect. P(kth symbol is transmitted correctly)=P(0 is transmitted and received correctly)+P(1 is transmitted and received correctly)=$$p(1-\epsilon_{0})+(1-p)(1-\epsilon_{1}) $$ b) P(1011 is...

A gas flows along an insulated pipe (q=0) through a porous plate that separates two sections of the pipe at different constant pressures P1 and P2. In Thermodynamics we study systems at equilibrium but here the gas is not in equilibrium because there is a pressure difference across the porous...

Wow I dont know why I was overcomplicating things. The probability that Bob wins= p+q*1/2=p+(1-2p)*1/2=1/2 My counting attempt was that every sequence with x heads for Bob can be paired with every Alice sequence having fewer than x heads. So the total number of pairs for this fixed x is $$...

Whether Bob has more heads depends on both Bob and Alice outcomes. For Bob to win Alice must get fewer heads. If bob gets x heads then Alice must get 0,1,2,..,x-1 heads. The number of sequences with k heads is $$ {n} \choose {k} $$. So for a fixed x the number of Alice sequences with less than x...

Z=Vreal gas/V ideal gas An Ideal gase assumes the only interaction between molecules is that they elastically bounce off each other it ignores attractive/repulsive force intermolecular forces (except during collisions). Does this mean that the ratio of intermolecular distance to molecular...

Appreciate everyones help.I didnt show path dependency. My question comes from recalling multivariable calculus, where the gravitational force is a conservative vector field and therefore its work is path independent. In the quasi-static piston setup, the external pressure is generated by...

Consider a gas contained in a rigid cylinder with a frictionless, weightless piston. The cylinder is in thermal contact with a thermostat at fixed temperature T. The space above the piston is evacuated. Initially, the piston is held by stops and the gas occupies a volume V1. If the gas is...

Does $$ (1-p(m,n-1)) $$ correspond to the sum of the probabilities that the first player wins odd numbered future turns conditioned on the fact that the first draw was black?

$$ P(\text{The event that the randomly chosen ball from jar 1 is white which is transferred to jar 2})=\frac{m}{m+n} $$ For the last transfer there are two disjoint events: Conditioned on the event that the ball transferred from jar k-1 to jar k is white the probability that the randomly chosen...

The recursive probability derived gives the probabilty the game ends on turn k and its valid when k>=2. The first player wins if the game ends when X is odd. $$P(\text{First player wins}) =\sum_{i=0}^{\frac{n}{2}} P(X=2i+1) $$ Since the game ends after all black balls are exhausted. $$ 2i+1...

Conditioning on this event the probability of winning on turn 2 given that the game was not won on the first turn is $$ P(X=2)= \frac{n}{m+n} \frac{m}{m+n-1} $$ $$ P(X=3)=\frac{n}{m+n}\frac{n-1}{m+n-1}\frac{m}{m+n-2} $$ If we generalize this argument $$ P(X=k)=P(X=k-1)\frac{n-k+2}{m+n-k+1} $$