I can't find solutions to these two and desperately need help :
20. An ideal diatomic gas is initially at temperature T and volume V. The gas is taken through three reversible processes in the following cycle: adiabatic expansion to the volume 2V, constant volume process to the temperature T, isothermal compression to the original volume V. For the complete cycle described above, which of the following is true?
(A) Net thermal energy is transferred from the gas to the surroundings.
(B) The net work done by the gas on the surroundings is positive.
(C) The net work done by the gas on the surroundings is zero.
(D) The internal energy of the gas increases.
(E) The internal energy of the gas decreases.
24. A counter near a longlived radioactive source measures an average of 100 counts per minute. The probability that more than 110 counts will be recorded in a given oneminute interval is most nearly
(A) zero (B) 0.001 (C) 0.025 (D) 0.15 (E) 0.5
ETS Extra #20 and #24  thermo and nuclear decay
Re: ETS Extra #20 and #24  thermo and nuclear decay
20. D and E are immediately out since the state of the gas is the same after a full cycle. Under adiabatic expansion the temperature of the gas drops. If you draw the PV diagram of the cycle you'll find the net work of the gas on the environment is negative (which means the thermal energy transferred to the environment is positive).
24. The rule of thumb for counting radioactive decays over a short amount of time relative to the half life is that it approaches a Poisson distribution and so the variance is approximately equal to the expected count. The expected count here is 100 so the standard deviation is sqrt(100) = 10. The question asks what the chance is that the number of counts is more than 1 standard deviation above the mean, which is about 15% since it's nearly a normal distribution.
24. The rule of thumb for counting radioactive decays over a short amount of time relative to the half life is that it approaches a Poisson distribution and so the variance is approximately equal to the expected count. The expected count here is 100 so the standard deviation is sqrt(100) = 10. The question asks what the chance is that the number of counts is more than 1 standard deviation above the mean, which is about 15% since it's nearly a normal distribution.

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Re: ETS Extra #20 and #24  thermo and nuclear decay
I think that you refer to the Gaussian distribution (which is used instead of the Poisson one according to the central limit theorem ...), Poisson distribution is used, in practice, only for very low counting rates. 100 counts per minute is >>1 and the Gaussian approximation does the job!kroner wrote:20. D and E are immediately out since the state of the gas is the same after a full cycle. Under adiabatic expansion the temperature of the gas drops. If you draw the PV diagram of the cycle you'll find the net work of the gas on the environment is negative (which means the thermal energy transferred to the environment is positive).
24. The rule of thumb for counting radioactive decays over a short amount of time relative to the half life is that it approaches a Poisson distribution and so the variance is approximately equal to the expected count. The expected count here is 100 so the standard deviation is sqrt(100) = 10. The question asks what the chance is that the number of counts is more than 1 standard deviation above the mean, which is about 15% since it's nearly a normal distribution.
Re: ETS Extra #20 and #24  thermo and nuclear decay
First of all, "normal distribution" is a synonym of "Gaussian distribution". But more importantly, just comparing the distribution to a Gaussian isn't enough to solve the problem. The variance of a Gaussian isn't determined by its expected value, so that alone doesn't give you enough information.
A key property of a Poisson distribution is that the variance is equal to the expected value, which is the first step of solving the problem. Alternately you could know that a binomial distribution for N particles each with a decay probability of p has an expected value of Np and a variance of Np(1p), so for p << 1 the variance is approximately equal to the mean.
A key property of a Poisson distribution is that the variance is equal to the expected value, which is the first step of solving the problem. Alternately you could know that a binomial distribution for N particles each with a decay probability of p has an expected value of Np and a variance of Np(1p), so for p << 1 the variance is approximately equal to the mean.
Re: ETS Extra #20 and #24  thermo and nuclear decay
Kroner  thanks for your help!