When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. `q = W = 500 J, Delta U = 0`B. `1 = Delta U = 500 J, W = 0`C. `q = W = 500 J, Delta U = 500`D. `Delta U = 0, q = W = - 500 J`

When 1 mole of gas is heated at constant volume. Temperature is raised

1.	When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. `q = W = 500 J, Delta U = 0`B. `1 = Delta U = 500 J, W = 0`C. `q = W = 500 J, Delta U = 500`D. `Delta U = 0, q = W = - 500 J`
Answer» Correct Answer - B According to the first law of thermodynamics, `Delta U = q + W` Since `W = - P_("ext") Delta V)` we have `Delta U = q + (- P_("ext") Delta V)` At constant volume, `Delta V = 0`. Thus, `Delta U = q` Since heat is supplied to the system, `q` is positive. Thus, `Delta U = q = 500 J, W = 0`

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When 1 mole of gas is heated at constant volume. Temperature is raised from `298` to `308K` . Heat supplied to the gas is `500J` . Then which stamenet is correct?A. `q = W = 500 J, Delta U = 0`B. `1 = Delta U = 500 J, W = 0`C. `q = W = 500 J, Delta U = 500`D. `Delta U = 0, q = W = - 500 J`

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