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?

Options

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

B

Explanation

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`