A2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final volume v2. a total of 1.7 kj of heat is added during the expansion process. what is v2? let the ideal-gas constant r = 8.314 j/(mol • k). a- 41 l b- 26 l c- 32 l d- 35 l

Isothermal Work =  PVln(v₂/v₁)

PV = nRT =  2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work =  PVln(v₂/v₁)            v₂ = ? v₁ = 19L, 

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) =

v₂  =  19*

v₂ = 25.8999

v₂ ≈ 26 L        Option b.

1. B. Increase

2. A. Decrease

Explanation:

To understand this issue, we need to put some values and using the ohm's law we can corroborate the two situations.

Ohm's law:

Now for the voltage we will use V = 110 [V], for resistance R = 10 [ohms]

Replacing the values we have:

Now let's double the voltage 220 [V]:

Therefore the current will be increased.

Let's do the same for the resistance if originally we have R = 10 [ohms]

Now let's double the resistance 20 [ohms]:

Therefore the current will be decreased.