Given: - Work done by the gas during isobaric expansion: \(W = 200 \, \text{J}\) - For a diatomic gas, the ratio of specific heats \(\gamma = 1.4\).
Step 1: Relationship for an Isobaric Process
In an isobaric process, the heat supplied \(Q\) to the system is given by:
\[ Q = \Delta U + W \]
where \(\Delta U\) is the change in internal energy of the gas and \(W\) is the work done by the gas.
Step 2: Change in Internal Energy
The change in internal energy for a diatomic gas is given by:
\[ \Delta U = nC_V\Delta T \]
For a diatomic gas, the molar specific heat at constant volume \(C_V\) is:
\[ C_V = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{5R}{2} \]
The molar specific heat at constant pressure \(C_P\) is given by:
\[ C_P = C_V + R = \frac{5R}{2} + R = \frac{7R}{2} \]
Thus, for an isobaric process, the heat \(Q\) is given by:
\[ Q = nC_P\Delta T = \frac{7}{5}\Delta U \]
Using the relation between work and internal energy change for an isobaric process:
\[ W = \frac{2}{5}Q \]
Substituting the given value of \(W\):
\[ 200 = \frac{2}{5}Q \]
Solving for \(Q\):
\[ Q = \frac{5}{2} \times 200 = 500 \, \text{J} \]
Conclusion: The heat given to the gas during the process is \(700 \, \text{J}\).
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: