Question

Calculate ∆U for each of the following cases:

Q = + 53 kJ, W = - 15 kJ
Q = + 100. kJ, W = - 62 kJ
Q = - 65 kJ, W = - 30 kJ
In which of these cases does the system do work on the surroundings?

Answer 1

Answer:

Q=+100kj,w=-15kj,Q=100kj,w=-62kj

Explanation:

when energy is exerted into a system work done is equal to zero .hence the system does work to the surrounding.

Answer 2

$\bold{\huge{\red{\underline{ Solution }}}}$

Here, We have to,

• Calucate change in internal energy ΔU for the following cases

Case 1 :-

We know that,

Change in internal energy = heat required + Work done

That is,

$\bold{\pink{ ΔU = Q + W }}$

Here, we have,

• $\sf{Q ( heat\: required) = + 53 kJ}$
• $\sf{W ( Work\: done) = - 15 kJ}$

[ Here, + sign of Q denotes that heat is absorbed in the given chemical reaction and - sign of W denotes work done by the system ]

Subsitute the required values in the given formula :-

$\sf{ ΔU = + 53 + ( -15) }$

$\sf{ ΔU = 53 - 15 }$

$\sf{ ΔU = 38kJ }$

Hence, The change in internal energy is 38kJ and work done is by the system on the surroundings

Case 2 :-

We know that,

$\bold{\pink{ ΔU = Q + W }}$

Here, we have

• $\sf{Q ( heat\: required) = + 100 kJ}$
• $\sf{W ( Work\: done) = - 62 kJ}$

[ Here, + sign of Q denotes that heat is absorbed in the given chemical reaction and - sign of W denotes work done by the system ]

Subsitute the required values in the above formula :-

$\sf{ ΔU = + 100 + ( -62) }$

$\sf{ ΔU = 100 - 62 }$

$\sf{ ΔU = 38kJ }$

Hence, The change in internal energy is 38kJ and work done is by the system on the surroundings .

Case 3 :-

We know that,

$\bold{\pink{ ΔU = Q + W }}$

Here, we have

• $\sf{Q ( heat\: required) = - 65 kJ}$
• $\sf{W ( Work\: done) = - 30 kJ}$

[ Here, - sign of Q denotes that heat is evolved in the given chemical reaction and - sign of W denotes work done by the system ]

Subsitute the required values in the above formula :-

$\sf{ ΔU = - 65 +(-30) }$

$\sf{ ΔU = -65 - 30 }$

$\sf{ ΔU = - 95kJ }$

Hence, The change in internal energy is -95 kJ and work done is by the system on the surroundings .

In which of this cases does the system do work on the surroundings?

• From above answer, we can conclude that ,

In all the three cases the work done is done by the system that is work is done on the surroundings by the system because work done is negative in all the three cases

Whereas, If work done is positive then work is done on the system not by the system.