Answer:
Explanation:
We know that the differential work made by the gas its defined as:
[tex]dW = P \ dv[/tex]
We can solve this by integration:
[tex]\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv[/tex]
but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law
[tex]P \ V = \ n \ R \ T[/tex]
[tex]P = \frac{\ n \ R \ T}{V}[/tex]
This give us
[tex] \int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv [/tex]
As n, R and T are constants
[tex] \int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv [/tex]
[tex] \Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1} [/tex]
[tex] \Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 ) [/tex]
[tex] \Delta W = \ n \ R \ T ( ln (v_2) - ln (v_1 ) [/tex]
[tex] \Delta W = \ n \ R \ T ln (\frac{v_2}{v_1})[/tex]
But the volume is:
[tex]V = \frac{\ n \ R \ T}{P}[/tex]
[tex] \Delta W = \ n \ R \ T ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )[/tex]
[tex] \Delta W = \ n \ R \ T ln(\frac{P_1}{P_2})[/tex]
Now, lets use the value from the problem.
The temperature its:
[tex]T = 27 \° C = 300.15 \ K[/tex]
The ideal gas constant:
[tex]R = 8.314 \frac{m^3 \ Pa}{K \ mol}[/tex]
So:
[tex] \Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K ln (\frac{20 atm}{1 atm}) [/tex]
[tex] \Delta W = 7475.69 joules[/tex]
We know that, for an ideal gas, the energy is:
[tex]E= c_v n R T[/tex]
where [tex]c_v[/tex] its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know
[tex]\Delta E = \Delta Q - \Delta W[/tex]
where [tex]\Delta W[/tex] is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:
[tex]\Delta E = 0[/tex]
[tex]\Delta Q = \Delta W[/tex]
