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One mole of an ideal gas at an initial temperature true of `TK` does `6R` joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is `5//3`, the final temperature of the gas will beA. `(T+2.4)K`B. `(T-2.4)K`C. `(T+4)K`D. `(T-4)K` |
Answer» Correct Answer - D `W_("adiabatic") = (nR(T_(1)-T_(2)))/(gamma-1) = 6R` `implies (1xxR(T_(1)-T_(2)))/(5//3-1) = 6R implies T_(2) = (T-4)K` |
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