Question

hello please help me with this ...

Answer 1

b) answer

w= - Pext ΔV = -2.5( 4.50 - 2.50 )
 = - 5L atm = - 5 × 101.325 J = - 506.625 J
Δ U = q + w

As, the container is insulated , thus q = 0
Hence , Δ U = w = - 506.625 J

Answer 2

Q:A gas is allowed to expand in a well insulated container against a constant external pressure of 2.t atm from an initial volume of 2.50 L to a final volume of 4.50 L.The change in internal energy ◇U of the gas in Joules will be

(a)-500 J ❎ (b)-505 J ✔

(c)+505 J ❎ (d)1136.25 J❎


Answer▪▪b)-505 J

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