When an ideal diatomic gas is heated at constant pressure , its internal energy increased by 50cal then  the work done by the gas is ___cal

To find the work done by the gas, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

Given that the internal energy of the gas increased by 50 cal, we can write this as:

ΔU = 50 cal

Since the gas is heated at constant pressure, we can use the equation:

ΔU = q - PΔV

Where q is the heat added to the system, P is the pressure, and ΔV is the change in volume.

Since the pressure is constant, we can rewrite the equation as:

ΔU = q - PΔV = q - P(Vf - Vi)

Where Vf is the final volume and Vi is the initial volume.

Since the gas is diatomic, we know that the change in volume is related to the change in temperature by the equation:

ΔV = αVΔT

Where α is the coefficient of volume expansion and ΔT is the change in temperature.

Since the gas is ideal, we can use the equation of state:

PV = nRT

Where n is the number of moles of gas and R is the ideal gas constant.

Combining these equations, we can write:

ΔU = q - P(Vf - Vi) = q - nRΔT

Given that the pressure is constant, we can rewrite the equation as:

ΔU = q - P(Vf - Vi) = q - nCpΔT

Where Cp is the molar heat capacity at constant pressure.

Now, we can substitute the given values into the equation:

ΔU = 50 cal Cp = ? ΔT = ?

Since we don't have the values for Cp and ΔT, we cannot calculate the exact work done by the gas. We would need more information to solve the problem.