Question

2.8g of $n_2$ gas at 300 k and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. calculate w for the

Explanation:

Step1: Calculate the number of moles of $N_2$

The molar mass of $N_2$ is $M = 28\ g/mol$. The number of moles $n=\frac{m}{M}$, where $m = 2.8\ g$. So $n=\frac{2.8\ g}{28\ g/mol}=0.1\ mol$.

Step2: Use the formula for work in an isothermal - irreversible expansion

The formula for work done in an isothermal irreversible expansion against a constant external pressure $P_{ext}$ is $W=-P_{ext}\Delta V$. For an ideal gas, $PV = nRT$. Initially, $P_1V_1=nRT$ and finally $P_2V_2=nRT$ (since isothermal, $T$ is constant). We can find the initial volume $V_1=\frac{nRT}{P_1}$, where $n = 0.1\ mol$, $R=0.0821\ L\cdot atm/(mol\cdot K)$, $T = 300\ K$, and $P_1 = 20\ atm$. So $V_1=\frac{0.1\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times300\ K}{20\ atm}=0.12315\ L$. The final volume $V_2=\frac{nRT}{P_2}$, with $P_2 = 1\ atm$. So $V_2=\frac{0.1\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times300\ K}{1\ atm}=2.463\ L$. The change in volume $\Delta V=V_2 - V_1=2.463\ L- 0.12315\ L = 2.33985\ L$.

Step3: Calculate the work

$W=-P_{ext}\Delta V$. Substituting $P_{ext}=1\ atm$ and $\Delta V = 2.33985\ L$, we get $W=-1\ atm\times2.33985\ L=- 2.34\ L\cdot atm$. Converting to joules, since $1\ L\cdot atm = 101.3\ J$, $W=-2.34\ L\cdot atm\times101.3\ J/(L\cdot atm)=-237\ J$.

Answer:

$-237\ J$