Here are some questions related to Radioactivity with their solutions:

Question 1:

8.5J heat flows out into the Surroundings when a sample of a gas contracts 400 mL by an average pressure of 0.5 atm. Calculate the change in energy of the system.

Solution :

Here, $\Delta V = -400 mL = -0.4L$ $and P = 0.5 atm.$ $w = P \Delta V$ $= -(0.5 atm)(-0.4 L)$ $0.2L atm$ $0.2 \times 101.2 J (\because 1 L atm = 101.2 J)$ $20.26J$ $\text{given that} q = -8.5 J (\because \text{Heat flows out of the system} )$ $\Delta E = q + w$ $-8.5 J + 2.0.26 J \\ 11.76 J$ $\Delta E$ has positive sign. It means energy of the system increases.

Questions 2:

10 moles of oxygen is kept in 5.0 L cylinder at 313 K. Due to sudden leakage through a hole, all the gas escapes into the atmosphere and the cylinder gets empty. If the atmospheric pressure is 1.0 atm, calculate the work done by the gas.

Solution : According to gas equation,

PV = nRT nRT $V = \dfrac{nRT}{P}$ $\dfrac{(10 ml)(0.082 L atm K^{-1} mol^{-1})(313K)}{1.0 atm}$ $= 256.66 L$ $\text{now} \Delta V = V_f- V_i$ $256.66 L- 5l$ $251.66 L$ $w = -P\Delta V$ $= -(1.0 atm)(251.66 L)$ $-251.66 L atm$ $-251.66 \times 101.3 J$ $-25493.2 J$ $= -25.4932kJ$

The energy change, $\Delta E$ for the reaction: $2N_2(g) + O_2(g) \to 2N_2O(g)$ is found to be $166 kJ at 25^0C$. Calculate the value of enthalpy change, $\Delta H$, at the same temperature.

Solution : For the reaction $2N_2(g) + O_2(g) \to 2N_2O(g)$ $\Delta n = 2- (2 + 1) = -1 \text{mole}$

According to gas equation, $P \Delta V = \Delta nRT \\[3mm] = (-1 mol)(8.314 JK^{-1} mol^{-1})(298 K) \\[3mm] -2477.6J \\[3mm] = -2.477 kJ$

We know that, $\Delta H = \Delta E + P \Delta V \\[3mm] = 166kJ + (-2.477 kJ) \\[3mm] 163.523 kJ \hspace{2mm}mol^{-1}$

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