10 gm of oxygen at a pressure `3 xx 10^5 N//m^2` and temperature `10^@C` is heated at constant pressure and after heating it occupies a volume of 10 litres. (a) Find the amount of heat received by the gas and (b) the energy of thermal motion of gas molecules before heating.

Answer

To solve the problem step by step, let's break it down into two parts as per the question.

### Part (a): Find the amount of heat received by the gas.

1. **Identify the given data:**
- Mass of oxygen, \( m = 10 \, \text{g} = 0.01 \, \text{kg} \)
- Pressure, \( P = 3 \times 10^5 \, \text{N/m}^2 \)
- Initial temperature, \( T_1 = 10^\circ C = 283 \, \text{K} \) (Convert to Kelvin by adding 273)
- Final volume, \( V = 10 \, \text{liters} = 0.01 \, \text{m}^3 \)

2. **Calculate the number of moles of oxygen:**
- Molar mass of oxygen, \( M = 32 \, \text{g/mol} \)
- Number of moles, \( N = \frac{m}{M} = \frac{10 \, \text{g}}{32 \, \text{g/mol}} = 0.3125 \, \text{mol} \)

3. **Use the ideal gas law to find the final temperature:**
- The ideal gas equation is \( PV = nRT \)
- Rearranging gives \( T_2 = \frac{PV}{nR} \)
- Using \( R = 8.314 \, \text{J/(mol K)} \):
\[ T_2 = \frac{(3 \times 10^5 \, \text{N/m}^2)(0.01 \, \text{m}^3)}{0.3125 \, \text{mol} \times 8.314 \, \text{J/(mol K)}} \]
- Calculate \( T_2 \):
\[ T_2 = \frac{3000}{2.604375} \approx 115.1 \, \text{K} \approx 585.65 \, \text{K} \]

4. **Calculate the change in temperature:**
\[ \Delta T = T_2 - T_1 = 585.65 \, \text{K} - 283 \, \text{K} = 302.65 \, \text{K} \]

5. **Calculate the heat received by the gas at constant pressure:**
- For an ideal gas, the heat absorbed at constant pressure is given by:
\[ Q = N C_P \Delta T \]
- Where \( C_P = \frac{7}{2} R \):
\[ C_P = \frac{7}{2} \times 8.314 \approx 29.19 \, \text{J/(mol K)} \]
- Now substituting the values:
\[ Q = 0.3125 \, \text{mol} \times 29.19 \, \text{J/(mol K)} \times 302.65 \, \text{K} \]
- Calculate \( Q \):
\[ Q \approx 0.3125 \times 29.19 \times 302.65 \approx 54.2 \, \text{J} \]

### Part (b): Find the energy of thermal motion of gas molecules before heating.

1. **Calculate the internal energy before heating:**
- The internal energy \( U \) of an ideal gas is given by:
\[ U = N C_V T \]
- Where \( C_V = \frac{5}{2} R \):
\[ C_V = \frac{5}{2} \times 8.314 \approx 20.79 \, \text{J/(mol K)} \]
- Now substituting the values:
\[ U = 0.3125 \, \text{mol} \times 20.79 \, \text{J/(mol K)} \times 283 \, \text{K} \]
- Calculate \( U \):
\[ U \approx 0.3125 \times 20.79 \times 283 \approx 38.7 \, \text{J} \]

### Final Answers:
(a) The amount of heat received by the gas is approximately **54.2 J**.
(b) The energy of thermal motion of gas molecules before heating is approximately **38.7 J**.