7. 1 molar ideal gas, its cv.m = 2.5r, underwent its temperature decrease of 50k at constant volume, please…

Explanation:

Step1: Calculate work done W

At constant volume, $dV = 0$. According to the formula $W=-P\Delta V$, when $\Delta V = 0$, $W = 0$. $W=0$

Step2: Calculate internal - energy change $\Delta U$

For an ideal gas, $\Delta U = nC_{V,m}\Delta T$. Given $n = 1mol$, $C_{V,m}=2.5R$, and $\Delta T=- 50K$. $\Delta U=nC_{V,m}\Delta T=1mol\times2.5R\times(-50K)=-125R$

Step3: Calculate heat transfer Q

According to the first - law of thermodynamics $\Delta U = Q+W$. Since $W = 0$, then $Q=\Delta U$. $Q=-125R$

Step4: Calculate enthalpy change $\Delta H$

For an ideal gas, $C_{p,m}=C_{V,m}+R = 2.5R+R = 3.5R$, and $\Delta H=nC_{p,m}\Delta T$. $\Delta H=nC_{p,m}\Delta T=1mol\times3.5R\times(-50K)=-175R$

Answer:

$W = 0$; $Q=-125R$; $\Delta U=-125R$; $\Delta H=-175R$