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.