The specific heat of argon at constant volume is `0.075 kcal kg^(-1) K^(-1)`. Calculate its atomic weight. `Take R = 2 cal mol^(-1) K^(-1)`.

Explanation

As argon is monoatomic ,its molar specific heat at constant volume will be
`C_(V) = (3)/(2)R = (3)/(2) xx 2 = 3 "cal" // "mol"K, C_(V) = M_(w)c_(v)` `" "` and `c_(v)= 0.075"cal"//gK`
So `3 = M_(w) xx 0.075 implies M_(w) = (3)/(0.075) = 40"gram"//"mole"`
Effecieny of a cycle `(eta)`:
`eta = ("total Mechanical work done by the gas in the whole process")/("Heat absorbed by the gas (only +ve)") = ("area under the cycle in " P-V " curve")/("Heat injected into the system")`