Solutions for Radiactivity Questions





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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