The temperature of a fixed mass of an ideal gas goes from 40ºC to 80ºC at a constant volume. If the initial pressure was x, what is the new pressure?

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The temperature of a fixed mass of an ideal gas goes from 40ºC to 80ºC at a constant volume. If the initial pressure was x, what is the new pressure?

The mark scheme says it's 2x, but do we use ºC or K in the equation P1/T1 = P2/T2?

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Answer 1 · 2017-12-21T17:41:54

Celsius.

Explanation:

Given:

$P_{1} = x$ $P_1=x$
$T_{1} = 40^{o} C = 313 K$ $T_1=40^oC=313K$
$T_{2} = 80^{o} C = 353 K$ $T_2=80^oC=353K$

We can see that the temperature is not the same ratio in Kelvin as Celsius, as while 40 is half of 80, 313 is not half of 353.

With Kelvin:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1=P_2/T_2$

$\Rightarrow P_{2} = \frac{P_{1} T_{2}}{T_{1}}$ $=>P_2=(P_1T_2)/(T_1)$

$\Rightarrow P_{2} = x \cdot \frac{353}{313}$ $=>P_2=x*353/313$

With Celsius:

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $P_1/T_1=P_2/T_2$

$\Rightarrow P_{2} = \frac{P_{1} T_{2}}{T_{1}}$ $=>P_2=(P_1T_2)/(T_1)$

$\Rightarrow P_{2} = x \cdot \frac{80}{40}$ $=>P_2=x*80/40$

$\Rightarrow P_{2} = 2 x$ $=>P_2=2x$

So it appears they wanted you to use Celsius.

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The temperature of a fixed mass of an ideal gas goes from 40ºC to 80ºC at a constant volume. If the initial pressure was x, what is the new pressure?

The mark scheme says it's 2x, but do we use ºC or K in the equation P1/T1 = P2/T2?

1 Answer

Explanation:

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