Question

Shows a cylindrical container containing oxygen \[\left( {\gamma = 1.4} \right)\] and closed by a 50Kg frictionless piston. The area of the cross section is $100c{m^2}$, atmospheric pressure in 100k pa and g is $10m{s^{ - 2}}$. The cylinder is slowly heated for some time. Find the amount of heat supplies to the gas if the piston moves out through a distance of 20cm.

Answer 1

Hint: Here, the pressure of the system will remain constant as the heat is supplied to the container the piston moves up 20cm and hence the pressure will remain constant. Make a relation between the work done, ideal gas equation and formula for heat capacity.

Complete step by step solution:
Here the work done by the gas:
$W = PdV$;
Here the pressure will be the initial pressure ${P_o}$and the pressure that is in the form of force upon area.
$ \Rightarrow W = \left( {{P_o} + \dfrac{{mg}}{A}} \right)Adx$;
We have been given the initial pressure is 100Kpa put this and the rest of the given value in the above equation:
$W = \left( {{{10}^5} + \dfrac{{50 \times 10}}{{100 \times {{10}^{ - 4}}}}} \right)\left( {100 \times {{10}^{ - 4}}} \right)\left( {20 \times {{10}^{ - 2}}} \right)$;
$ \Rightarrow W = \left( {{{10}^5} + \dfrac{{500}}{{{{10}^{ - 2}}}}} \right)\left( {20 \times {{10}^{ - 4}}} \right)$;
Do the needed calculation:
\[ \Rightarrow W = \left( {\dfrac{{{{10}^3} + 500}}{{{{10}^{ - 2}}}}} \right)\left( {20 \times {{10}^{ - 4}}} \right)\];
\[ \Rightarrow W = \left( {1500} \right)\left( {20 \times {{10}^{ - 2}}} \right)\];
The work done by the gas is:
\[ \Rightarrow W = 300J\];
We have the ideal gas equation as:
$PV = nRT$;
Now, differentiate on both the sides:
$PdV + VdP = nRdT$;
Here there is no change in pressure as the pressure will remain same in the initial and final conditions, the term with differential pressure will get zero:
$ \Rightarrow PdV = nRdT$;
Now we know that the work done by a gas equals pressure times the change in volume “$W = PdV$”
$nRdT = W$;
We know that the work done is 300J so, put the value of work in the above relation:
$nRdT = 300$;
Solve for dT:
$ \Rightarrow dT = \dfrac{{300}}{{nR}}$;
The heat capacity is defined as:
$Q = n{C_p}dT$;
Put the value of dT in the above equation of heat capacity:
$\Rightarrow Q = n{C_p}\dfrac{{300}}{{nR}}$;
Now, the gas constant “R” is given as the subtraction of specific heat constant at pressure by specific heat constant at volume:
$R = {C_p} - {C_v}$;
Put the above relation in the equation “$Q = n{C_p}\dfrac{{300}}{{nR}}$”:
$\Rightarrow$ \[Q = \dfrac{{{C_p} \times 300}}{{{C_p} - {C_v}}}\];
Now, Divide the numerator and denominator by\[{C_p}\]:
$\Rightarrow$ \[Q = \dfrac{{300}}{{1 - \dfrac{{{C_v}}}{{{C_p}}}}}\];
The ratio of heat capacities \[\dfrac{{{C_v}}}{{{C_p}}}\]is given as:
$\Rightarrow$ \[\dfrac{1}{\gamma } = \dfrac{{{C_v}}}{{{C_p}}}\];
Now, put this in the above equation:
\[\Rightarrow Q = \dfrac{{300}}{{1 - \dfrac{1}{\gamma }}}\]
\[ \Rightarrow Q = \dfrac{{300\gamma }}{{\gamma - 1}}\];
Now, we have been given the value “\[\left( {\gamma = 1.4} \right)\]”;
\[ \Rightarrow Q = \dfrac{{420}}{{0.4}}\];
\[ \Rightarrow Q = 1050J\];

Therefore, the amount of heat supplied to the gas if the piston moves out through a distance of 20cm is 1050J.

Note: Here, we need to first find the work done by the gas by equating work equals pressure time change in volume. Here the pressure would be the atmospheric pressure and the pressure put by the piston. Now, equate the work done with the differential form of the ideal gas equation and put it in the heat capacity formula and find out heat.