Question

a) what will be the temperature change if a (1255 j) amount of heat is added to a mono - atomic gas container at room temperature and at constant volume and contains n = 5.0 mole of gas?
b) to get same amount of temperature change of a diatomic gas container of n = 5.0 mole and at temperature t (700.0 k), how much energy q is needed at constant volume?
c) in part c diatomic case, estimate how much energy goes into rotational motion of gas molecule? neglect the vibrational degrees of freedom.
a) + q
1 2
δv = 0
v1 = v2
f = 3
p1 v1 t1
p2 v2 t2
δt =? = t2 - t1
q = ncvδt
δt = q / ncv = q / n(3/2)r = 2q / nfr = 2q / 3nr
cv = (3/2)r
δt = 2×1255 / 3×5.0×8.31 =? k
b) d =? f = 5 so d = ncvδt
@ 700.0k
d = n(5/2)rδt = 5/2 nrδt
= 5/2×5.0×8.31×?
?
c) q(rot) =?
q = q(trans) + q(rot)
if q(trans) = 1255 j.
q(rot) = q - 1255 j.
part(b)
=? j.

Explanation:

Step1: Recall the molar - specific heat at constant volume for a mono - atomic gas

For a mono - atomic gas, the molar - specific heat at constant volume $C_V=\frac{3}{2}R$, where $R = 8.31\ J/(mol\cdot K)$ is the universal gas constant. The heat added to the gas is given by $Q = nC_V\Delta T$, where $n$ is the number of moles of the gas. We are given $Q = 1255\ J$ and $n = 5.0\ mol$.

Step2: Solve for $\Delta T$

Rearranging the formula $Q=nC_V\Delta T$ for $\Delta T$, we get $\Delta T=\frac{Q}{nC_V}$. Substituting $C_V=\frac{3}{2}R$ into the formula, we have $\Delta T=\frac{Q}{n\times\frac{3}{2}R}$. Plugging in $Q = 1255\ J$, $n = 5.0\ mol$, and $R=8.31\ J/(mol\cdot K)$:

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Step3: Recall the molar - specific heat at constant volume for a diatomic gas

For a diatomic gas, the molar - specific heat at constant volume $C_V=\frac{5}{2}R$. The heat added to the gas at constant volume is $Q = nC_V\Delta T$.

Step4: Calculate the energy for a diatomic gas

We know $n = 5.0\ mol$, $C_V=\frac{5}{2}R$, and assume we want to find the heat added for a certain $\Delta T$. But if we consider the general formula $Q=nC_V\Delta T$, for a diatomic gas with $n = 5.0\ mol$, $Q = 5.0\ mol\times\frac{5}{2}\times8.31\ J/(mol\cdot K)\times\Delta T$.

Step5: Analyze the degrees of freedom for a diatomic gas

For a diatomic gas, the total internal energy is associated with translational and rotational degrees of freedom. The translational degrees of freedom $f_{trans}=3$ and the rotational degrees of freedom $f_{rot}=2$. The molar - specific heat for translational motion $C_{V,trans}=\frac{3}{2}R$ and for rotational motion $C_{V,rot}=\frac{2}{2}R = R$.
If the total heat added $Q$ and the heat for translational motion $Q_{trans}=1255\ J$, and since $Q_{trans}=n\times\frac{3}{2}R\Delta T$ and $Q_{rot}=n\times R\Delta T$.
We know that $\frac{Q_{rot}}{Q_{trans}}=\frac{nR\Delta T}{n\times\frac{3}{2}R\Delta T}=\frac{2}{3}$. So $Q_{rot}=\frac{2}{3}Q_{trans}$.
Since $Q_{trans}=1255\ J$, then $Q_{rot}=\frac{2}{3}\times1255\ J\approx837\ J$

Answer:

a) $\Delta T\approx20.1\ K$
b) (No specific $\Delta T$ given in the problem for calculating $Q$ for diatomic gas in a non - relative way. If we assume the same $\Delta T$ as in part a, $Q = nC_V\Delta T=5.0\ mol\times\frac{5}{2}\times8.31\ J/(mol\cdot K)\times20.1\ K\approx4180\ J$)
c) $Q_{rot}\approx837\ J$