Question

Dry ice is solid carbon dioxide. A $ 0.05\;g $ sample of dry ice is placed in an evacuated $ 4.6\;L $ vessel at $ {30^ \circ }C $ . Calculate the pressure inside the vessel after all the dry ice has been converted to $ C{O_2} $ gas.
(A) $ 6.14\;atm $
(B) $ 0.614\;atm $
(C) $ 0.0614\;atm $
(D) $ 6.14 \times {10^{ - 3}}atm $

Answer 1

Hint :We will solve this problem by using the ideal gas equation which is $ PV = nRT $ . Here we have to calculate the pressure inside the vessel. All of the values are given except for the number of moles. Let’s find out the solution stepwise.

Complete Step By Step Answer:
Dry ice is the solid formula of carbon dioxide which is primarily used as a cooling agent. You must have seen in movies or dramas where a white smoke is released to create dramatic effect. It is nothing but dry ice which is used in fog machines. It is also used in preserving frozen foods where mechanical cooling is not available.
We all know the ideal gas equation which is
 $ PV = nRT $
Where,
P is the pressure
 V is the volume
 n is the number of moles
 R is the universal gas constant
 T stands for temperature.
First of all we will find out the number of moles which is
Number of moles is given by dividing the given mass of a compound by its molar mass.
Here the compound is $ C{O_2} $ and the mass of $ C{O_2} $ is $ 44\;g $
Now number of moles of $ C{O_2} $ is
 $ n = \dfrac{{0.05}}{{44}} $
 $ = 0.0013mol $
Given:
 $ R = {\text{ }}0.0821 $
 $ V = 4.6L $
 $ T = {30^ \circ }C= 303K $
Now putting all these value in the ideal gas equation
 $ PV = nRT $
 $ P = {\text{ }}\dfrac{{{\text{nRT}}}}{V} $
 $ P = {\text{ }}\dfrac{{0.001136 \times 0.0821 \times 303}}{{4.6}} $
 $ = 0.00614 $
 $ = 6.14 \times {10^{ - 3}}atm $
So our answer is option number D.

Note :
The ideal gas equation is used only for the ideal gas and cannot be applied to the real gases. The reason for this is due to certain assumptions which are taken into consideration for the ideal gas and cannot be true for the real gas, for example volume of the real gas cannot be negligible. So the answer is only a close value of the degree of dissociation.