`C_(P) - C_(V) = R`. This `R` is

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

The value of C_(P) - C_(v) = 1.00 R for a gas in state A and C_(P) - C_(v) = 1.06 R in another state. If P_(A) and P_(B) denote the pressure and T_(A) and T_(B) denote the temperature in the two states, then

The value of C_(P) - C_(v) = 1.00 R for a gas in state A and C_(P) - C_(v) = 1.06 R in another state. If P_(A) and P_(B) denote the pressure and T_(A) and T_(B) denote the temperature in the two states, then

For a gas C_(P)-C_(V)=R in a state P and C_(P)-C_(V)=1.10 R in a state Q, T_(P) and T_(Q) are the temperatures in two different states P and Q respectively. Then

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.

In Mayer's relation: C_(P)-C_(V)=R 'R' stands for:

In Mayer's relation: C_(P)-C_(V)=R 'R' stands for:

Show that C_P- C_V = R