Question 513 Marks
State first law of thermodynamics. Why $C_p > C_v$? Prove $C_p - C_v = R$.
Answer
View full question & answer→First Law of Thermodynamics: It is based on the law of conservation of energy. The total heat energy change in any system is the sum of the internal energy change and the work done, i.e. dQ = dU + dW. where dU is the internal energy change and dW = PdV is the work done by/ on the system. $C_p$ is greater because under constant pressure process, the energy also does work. Poof of $C_p - C_v = R$ Suppose one mole of a gas is heated at constant volume so that its temperature rises by dT. Heat supplied = $1 × C_V \times dT = C_VsdT$ ...(i) Since the volume is constant, the gas will not perform external work in accordance with the first law of thermodynamics and the heat supplied will be just equal to the increase in the internal energy of the gas.
$\therefore dU = C_VdT$ ...(ii) Let the gas be heated at constant pressure to increase its temperature by dT, and dQ be the amount of heat supplied, therefore, $dQ = 1 \times C_P \times dT = C_PdT$ ...(iii) The heat supplied at a constant pressure increases the temperature by dT, hence increases its internal energy by $dU = C_vdT$ as well as enables the gas to perform work dW. dW = PDV ...(iv) From the first law of thermodynamics, we have, dQ = dU + dW Substituting the values, we get, $C_PdT = C_VdT + PdV$ But PV = RT (For one mole of an ideal gas) or PdV = RdT,
$\therefore\text{C}_\text{P}\text{dt}=\text{C}_\text{P}\text{dT}+\text{RdT} C_P - C_V = R$ ...(v) This is the relationship between two principal specific heats of the gas when $C_p, C_v$ and R are measured in the units of either heat or of work.
$\therefore dU = C_VdT$ ...(ii) Let the gas be heated at constant pressure to increase its temperature by dT, and dQ be the amount of heat supplied, therefore, $dQ = 1 \times C_P \times dT = C_PdT$ ...(iii) The heat supplied at a constant pressure increases the temperature by dT, hence increases its internal energy by $dU = C_vdT$ as well as enables the gas to perform work dW. dW = PDV ...(iv) From the first law of thermodynamics, we have, dQ = dU + dW Substituting the values, we get, $C_PdT = C_VdT + PdV$ But PV = RT (For one mole of an ideal gas) or PdV = RdT,
$\therefore\text{C}_\text{P}\text{dt}=\text{C}_\text{P}\text{dT}+\text{RdT} C_P - C_V = R$ ...(v) This is the relationship between two principal specific heats of the gas when $C_p, C_v$ and R are measured in the units of either heat or of work.
