Question

consider a n = 4.6 mol of helium balloon and it undergoes temperature increases of δt = 32 °c at constant pressure of p = 2.3 atm (1.0 atm = 1.01×10^5 pa) (isobaric process). as a result, the balloon expands. the helium is monoatomic ideal gas. (a) what is the change in the internal energy of the helium during the temperature increase? (b) how much work is done by the helium as it expands against the pressure of the surrounding water during the temperature increase? (c) heat absorbed by the gas during isobaric expansion? a) δe = (\frac{3}{2}) nrδt = (\frac{3}{2})×4.6×8.31×32 =? j. b) w = (int pdv) = pδv = p(v₂ - v₁) = p (\frac{nr}{p})δt = nrδt = 4.6×8.31×32 =? j. c) q = δe + w = (\frac{3}{2})nrδt + nrδt > 0 = (? +?) j ≈? j. a sample of m = 0.125 kg of xenon (mono - atomic gas) is contained in a

Explanation:

Step1: Calculate change in internal energy

Helium is a mono - atomic gas with $C_V=\frac{3}{2}R$. The formula for change in internal energy $\Delta E = nC_V\Delta T$. Given $n = 4.6\ mol$, $R=8.31\ J/(mol\cdot K)$ and $\Delta T = 32\ K$. So, $\Delta E=\frac{3}{2}nR\Delta T=\frac{3}{2}\times4.6\times8.31\times32$.

$$
LATEXBLOCK0
$$

Step2: Calculate work done

For an isobaric process, $W = P\Delta V$. From the ideal gas law $PV = nRT$, so $\Delta V=\frac{nR\Delta T}{P}$. Then $W = nR\Delta T$. Substituting $n = 4.6\ mol$, $R = 8.31\ J/(mol\cdot K)$ and $\Delta T=32\ K$, we get $W=4.6\times8.31\times32$.

$$
LATEXBLOCK1
$$

Step3: Calculate heat absorbed

By the first law of thermodynamics $Q=\Delta E+W$. We know $\Delta E = 1834.848\ J$ and $W = 1223.232\ J$. So $Q=1834.848 + 1223.232=3058.08\ J$.

Answer:

(a) $\Delta E = 1834.848\ J$
(b) $W = 1223.232\ J$
(c) $Q = 3058.08\ J$