Question

A gas absorbs 500J heat and utilizes Q J in doing work against an external pressure of 2atm. If $\Delta H$ is -510J, values of $\Delta V$ & W respectively are:
(A) 10$c{m^3}$ ; 500J
(B) ${10^3}c{m^3}$ ; -510J
(C) 10$d{m^3}$; 1010J
(D) $5d{m^3}$ ; -1010J

Answer 1

Hint: Heat and the work is related to each other by following formula
\[\Delta U = q + w\]
Work and pressure is related as
\[W = - P \cdot \Delta V\]

Complete step by step solution:
We know that the relation between the internal energy and the work can be given by the first law of thermodynamics.
So, this relation can be given as
\[\Delta U = q + w\]
Here, q is the heat and w is the work. We are given that $\Delta U = - 510J$. Here, heat (q) is given 500J.
- Heat absorbed at constant volume is measured practically by a Bomb calorimeter. It is a steel vessel immersed in a water bath.
So, we can put these values in the above equation as
\[ - 510 = 500 + w\]
So, we can write that
\[w = - 510 - 500 = - 1010J\]
Now, we obtained that the work (W) is -1010J.
Now, we will find the $\Delta V$ for the reaction. This can be obtained by the formula that relates work (W), external pressure and $\Delta V$. So, this formula can be given as
\[W = - P \cdot \Delta V\]
We are also given that external pressure is 2atm. So, we can write the above equation as
\[ - 1010 = - 2 \cdot \Delta V\]
So, we can write that
\[\Delta V = \dfrac{{1010J}}{{2atm}}\]
Now, we can write that $1L \cdot atm = 101.33J$
So, we can write that
\[\Delta V = \dfrac{{1010 \times 1L \cdot atm}}{{2atm \cdot 101.33}} = 5L\]
Thus, we obtained that $\Delta V$=5L.
Now, we can say that 1L=$1d{m^3}$
So, 5L=$5d{m^3}$.
Thus, we obtained that W is -1010J and $\Delta V$ is 5$d{m^3}$.

Therefore the correct answer is (D).

Note: Do not forget the negative sign in the equation of the work which relates to the pressure and the volume of the system. Remember that 1L=$1d{m^3}$. Here, the work is done by external pressure. So, it is negative in sign.