Answer:
The mass of propane gas which has been used in a barbecue is 195.05 g
Explanation:
Given
We are given a tank with length L =1 in and radius r = 0.06 m filled with a propane gas with pressure p = 1.3 x 10^6 Pa at temperature T = 22 C° = 295K and the volume V employed by
V=πr^2L
=π(0.06m)^2*1m
=0.0113m^3
Some of this gasemptied to be used and the pressure decreased to 2.50 x 10^5 Pa and definitely mass also reduced and we were asked to calculate the mass of gas that has been used.
Solution
In this case, we have two states: 1st state is propane gas has mass mi and pressure pi = 1.3 x 10^6 Pa
2nd state is propane gas has mass m2 remains in the tank after a part of the gas is emptied and pressure p2 = 2.50 x 10^5 Pa
We are given p1 and p2 and must calculate mi first to find m2 which will lead us to find the amount of the gas that has been used. So in first state we would apply Ideal gas law to find the initial mass m1
p1*V=nRT (1)
Where n is the number of moles and could sub with its value with m/M where Al is the molar mass of propane gas and equals 44.1 gimol as given. to find the relation between the number of moles, the mass and the molar mass) . So equation (1) would be
p1*V=(m1/M)*RT (2)
By sub with all given values in equation (2) and solve for mi
m1 = p1*V*M/RT
=264.13 g
By back to equation (2) we would find that the relation between pressure and mass is proportional (Note that the volume, the temperature, R and molar mass are constant). So we would extract a relation between p and m for the two states by
p1/p2=m1/m2
By sub with all values of p1, p2 and m1 and solve for m2
m2=m1*p2/p1
=69.08 g
Now we found the mass of remaining gas in the tank after a part of gas was used. lb calculate the mass of propane gas that has been used it would be given by
m1-m2 = 264.13 g - 69.08 g
=195.059 g
The mass of propane gas which has been used in a barbecue is 195.05 g
A cylinder 1.00m tall with inside diameter 0.120 m is used to hold propane at 22 °C and 1.30 × 10⁶ Pa when full and at 22 °C and 2.50 × 10⁵ Pa when partially emptied. The mass of propane consumed is 213 g.
A cylinder is 1.00 m tall (h) and has a diameter (d) of 0.120 m. We will calculate its volume (V) using the following expression.
[tex]V = h \times \pi \times (d/2)^{2} = 1.00 m \times \pi \times (0.120/2)^{2} = 0.0113 m^{3}[/tex]
Initially, the gas propane occupying the cylinder at 22 °C (295 K) has a pressure of 1.30 × 10⁶ Pa. We can calculate the moles of propane using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.30 \times 10^{6}Pa \times 0.0113m^{3} }{\frac{8.314Pa.m^{3} }{mol.K} \times 295K} = 5.99 mol[/tex]
After it is partially emptied, the pressure in the tank is 2.50 × 10⁵ Pa. We can calculate the moles of propane using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{2.50 \times 10^{5}Pa \times 0.0113m^{3} }{\frac{8.314Pa.m^{3} }{mol.K} \times 295K} = 1.15 mol[/tex]
Initially, there were 5.99 moles and finally 1.15 moles. The moles of propane used were:
[tex]5.99mol-1.15mol = 4.84 mol[/tex]
We can convert 4.84 moles to grams using the molar mass of propane (44.1 g/mol).
[tex]4.84 mol \times \frac{44.1g}{mol} = 213 g[/tex]
A cylinder 1.00m tall with inside diameter 0.120 m is used to hold propane at 22 °C and 1.30 × 10⁶ Pa when full and at 22 °C and 2.50 × 10⁵ Pa when partially emptied. The mass of propane consumed is 213 g.
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