Increased pressure of gas.

Classical Mechanics Level 3

There is a bound cylinder with some gas in it. The temperature of that gas is increased from $40^{\circ}C$ to $90^{\circ} C$ , then the pressure of it is increased by $\dfrac ab \%$ . [ $a,b \in \mathbb Z , gcd(a,b) =1$ ]

Find $(a+b)^{\circ} C$ in fahrenheit.

Details and assumptions:

The gas follows the ideal gas law .
$273 K =0^{\circ} C$

The answer is 9595.4.

1 solution

Munem Shahriar
Aug 21, 2018

Initial pressure, $P_1$ .
Changed pressure, $P_2$ .
Initial temperature, $T_1 = 40^{\circ} C = 40 +273 = 313 K$
Changed temperature, $T_2 = 90^{\circ} C = 90 + 273 = 363 K$

We know that, $PV = nRT$ [Ideal gas law]

So,

$P_1 V = nR T_1 ~~~~~....... (1)$

$P_2 V = nR T_2 ~~~~ .......(2)$

$(1) \div (2):$

$\begin{aligned} \dfrac{P_1 V}{P_2 V} & = \dfrac{nRT_1}{nR T_2} \\ \Rightarrow \dfrac{P_1}{P_2} & = \dfrac{T_1}{T_2} \\ \Rightarrow \dfrac{P_1}{P_2} & = \dfrac{313}{363} \\ \Rightarrow P_2 & = \dfrac{363 P_1}{313} \\ \end{aligned}$

Now, increased pressure of gas = $P_2 - P_1 = \dfrac{363 P_1}{313} - P_1 = \dfrac{50}{313} P_1$ . So $\dfrac ab \% = \dfrac{ 50}{313} \times 100 = \dfrac{5000}{313}\%$

So $(a + b)^\circ C = (5000 +313)^\circ C = 5313^\circ C$ .

We know that,

$\begin{aligned} \dfrac{T_C} 5 & = \dfrac{T_F - 32}9 \\ \Rightarrow \dfrac{5313}5 & =\dfrac{T_F - 32}9 ~~~ [T_C = 5313^\circ C] \\ \Rightarrow \dfrac{5313 \times 9} 5& = T_F - 32 \\ \Rightarrow 9563.4 +32 & =T_F \\ \implies T_F & = \boxed{9595.4^{\circ} F} \\ \end{aligned}$

5313 is a dimensionless number. It started as a ratio, expressed as a percent increase. Then artificially adding the numerator and denominator to get a single quantity (as is often done for an answer to these types of problems).

Attempting to put it into Centigrade, then converting to Fahrenheit does not make sense to me.

Steven Perkins - 2 years, 9 months ago

Increased pressure of gas.

The answer is 9595.4.

1 solution

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