Prove `C_(p) - C_(v) = R`

Prove that C_(p) - C_(v) = R , where C_(p) and C_(v) are the molar specific heats at constant pressure and constant volume respectively and R is the universal gas constant.

Show that for ideal gas C_(p)-C_(v) = R .

C_(p) - C_(v) = a for hydrogen gas and C_(p) - C_(v) = b for oxygen gas. What is the relation between a and b?

Show that for ideal gas C_(P ) - C_(v ) = R and gamma =1 +(2)/(f ) , where f is the degrees of freedom of the molecule .

C_(P) and C_(v) are specific heats at constant pressure and constant volume respectively. It is observed that C_(p) - C_(v) = a for hydrogen gas C_(p) - C_(v) = b" " for nitrogen gas The correct relation between a and b is

The molar specific heats C_(p) and C_(v) of an ideal gas are expressed in the unit of cal cdot mol^(-1) cdot^(@)C^(-1) . But R is expressed in J cdot mol^(-1) cdot^(@)C^(-1) . Write down the relation between C_(p) and C_(v) .

Find out the molar specific heats C_(p) an C_(v) of an ideal gas having gamma = 1.67 . Given, R = 2 cal cdot mol^(-1) cdot^(@)C^(-1) .

Define 1st law of thermodynamics. By using this law, derive relationship between C_(p) and C_(v) .

In case of 1 mol of an ideal gas, write down the value of C_(v) - C_(p) .