Question

the pressure of a certain amount of gas, in pascals (unit = pa) is modelled by the formula ( p=\frac{8.3(t + 273)}{v} ), where t represents the temperature of the gas, in ( { }^{circ} mathrm{c} ), and v represents the volume in ( mathrm{m}^{3} ).
a. determine the pressure of the gas, to 1 decimal, when the temperature is ( 24^{circ} mathrm{c} ) and the volume is ( 200 mathrm{~m}^{3} )
b. isolate the equation for t. if there is a fraction, use brackets to distinguish the numerator and denominator.
ex: ( \frac{3 x+y}{z + 2} ) should be written as ( (3 x+y) /(z + 2) )
determine the temperature of the gas, to 1 decimal, when the pressure is ( 36 mathrm{~pa} ) and the volume is ( 120 mathrm{~m}^{3} )

Explanation:

a.

Step1: Substitute values into formula

Given \(T = 24^{\circ}C\) and \(V=200m^{3}\), substitute into \(P=\frac{8.3(T + 273)}{V}\).

$$P=\frac{8.3(24+ 273)}{200}$$

Step2: Calculate numerator

First, calculate \(24 + 273=297\). Then \(8.3\times297 = 8.3\times(300-3)=8.3\times300-8.3\times3=2490 - 24.9=2465.1\)

Step3: Calculate \(P\)

Now, \(P=\frac{2465.1}{200}=12.3255\approx12.3\) (to 1 decimal place)

Step1: Start with \(P=\frac{8.3(T + 273)}{V}\)

Multiply both sides by \(V\): \(PV=8.3(T + 273)\)

Step2: Divide both sides by \(8.3\)

\(\frac{PV}{8.3}=T + 273\)

Step3: Subtract 273 from both sides

\(T=\frac{PV}{8.3}-273\)

Step1: Substitute \(P = 36\) and \(V = 120\) into \(T=\frac{PV}{8.3}-273\)

$$T=\frac{36\times120}{8.3}-273$$

Step2: Calculate \(\frac{36\times120}{8.3}\)

\(36\times120 = 4320\), then \(\frac{4320}{8.3}\approx520.5\)

Step3: Calculate \(T\)

\(T=520.5-273 = 247.5\)

Answer:

\(12.3\)

b.