500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K^-1 mol^-1

<p>Amount of energy transpired as hat, $Q=500 \mathrm{~J}$</p> <p>Number of moles of Argon, $n=0.5 \mathrm{~mol}$</p> <p>Temperature (Initial), $T_i=298 \mathrm{~K}$</p> <p>Pressure, $P=1.00 \mathrm{~atm}$</p> <p>Final temperature, $T_F=$ ?</p> <p>Change in internal energy, $\Delta V=$ ?</p> <p>$R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$</p> <p>For a constant pressure process, the heat added is related to the temperature changes as</p> <p>$Q=n C_p \Delta T$ ....... (1)</p> <p>$$\begin{aligned} &\text { Also, the change in internal energy formula is }\\ &\Delta U=n C_V \Delta T \quad-(2)\quad\text{.... (2)} \end{aligned}$$</p> <p>$$\begin{aligned} &\text { For monoatomic gas (Argon), }\\ &\begin{aligned} C_v & =\frac{3}{2} R \\ \text { So, } C_p & =C_v+R \quad\left(C_p-C_v=R\right) \\ & =\frac{3}{2} R+R=\frac{5}{2} R \end{aligned} \end{aligned}$$</p> <p>$$\begin{aligned} &\text { In equation (1) } \Rightarrow\\ &\begin{aligned} & Q=n C_p \Delta T \\ & Q=n C_p\left(T_F-T_i\right) \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} R\left(T_F-T_i\right) \quad R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ & 500 \mathrm{~J}=0.5 \mathrm{~mol} \times \frac{5}{2} \times 8.3 \mathrm{~J} K^{-1} \mathrm{~mol} ^{-1}\left(T_F-T_i\right) \end{aligned} \end{aligned}$$</p> <p>$T_F-T_i=\frac{500 \mathrm{~J}}{0.5 \mathrm{mol} \times \frac{5}{2} \times 8.3 \text { J K}^1 \mathrm{~mol}^{-1}}$</p> <p>$$\begin{aligned} & =\frac{500}{0.5 \times \frac{5}{2} \times 8.3} \mathrm{k} \\ & =48.2 \mathrm{k} \\ T_F & =48.2 \mathrm{k}+298 \mathrm{k} \\ & =346.2 \mathrm{k} \\ & \approx 348 \mathrm{k} \end{aligned}$$</p> <p>$$\begin{aligned} &\text { In equation (2) } \Rightarrow\\ &\begin{aligned} \Delta U & =n C_V \Delta T \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} R \times 48.2 \mathrm{~K} \\ & =0.5 \mathrm{~mol} \times \frac{3}{2} \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{mol^{-1}} \times 48.2 \mathrm{~K} \\ & =300.045 \mathrm{~J} \approx 300 \mathrm{~J} \end{aligned} \end{aligned}$$</p> <p>The final temperature is, 348 k</p> <p>change in internal energy is 300 J</p> <p>So, correct answer is Option 2) 348 K and 300 J</p>