Question

The pressure, volume, and temperature of a mole of an ideal gas are related by the equation PV = 8.31T, where P is measured in kilopascals, V in liters, and T in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from 10 L to 10.3 L and the temperature decreases from 330 K to 320 K. (Note whether the change is positive or negative in your answer. Round your answer to two decimal places.)

Answer 1

Answer:

The pressure will decrease by about 16.53 kPa.

Explanation:

Given that,

The equation is

$PV=8.31 T$...(I)

We need to calculate the initial pressure

Using given equation

$PV=8.31 T$

Put the value into the formula

$P=\dfrac{8.31\times330}{10}$

$P=274.23\ kPa$

The volume increase from 10 L to 10.3 L.

$dV=10.3-10=0.3\ L$

The temperature decrease from 330 K to 320 K.

$dT=320-330=-10\ K$

We need to calculate the change in the pressure

Now, on differentiating equation (I)

$d(PV)=8.31dT$

$PdV+VdP=8.31 dT$

Put the value into the formula

$274.23\times0.3+10\times dP=8.31\times(-10)$

$dP=\dfrac{-8.31\times10-274.23\times0.3}{10}$

$dP=−16.53\ kPa$

Hence, The pressure will decrease by about 16.53 kPa.

Answer 2

The approximate change in the pressure is -16.53 Pa

Gas equation:

The given gas equation is:

PV = 8.31T

where P is the pressure, V is the volume and T is the temperature

now if we take the differential of the above equation we get:

PdV + VdP =8.31dT

It is given that the change in volume is:

dV = 10.3L -10 L = 0.3 L

and the change in temperature is :

dT = 320k - 330k = -10k

According to the question, initially T = 330K and V = 10L

So, P = 8.31×330/10

P = 0.273 kPa

Now,

PdV + VdP =8.31dT

0.273(0.3) + 10(dP) = 8.31(-10)

dP = -16.53 Pa

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