Question

The pressure, volume, and temperature of a mole of an ideal gas are related by the equation PV ???? 8.31T, where P is mea- sured in kilopascals, V in liters, and T in kelvins. Use differ- entials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K.

Answer 1

Answer:3.38 kilopascals.

Step-by-step explanation:To find the approximate change in pressure when the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K, we can use differentials.

1. First, let's differentiate the given equation PV = 8.31T with respect to each variable:

- d(PV) = d(8.31T)

- P*dV + V*dP = 8.31*dT

2. We are interested in finding the change in pressure, so we'll focus on dP:

- P*dV + V*dP = 8.31*dT

- V*dP = 8.31*dT - P*dV

3. We can now substitute the specific values provided:

- V = 12.3 L - 12 L = 0.3 L (change in volume)

- T = 305 K - 310 K = -5 K (change in temperature)

4. Substituting these values into the equation, we get:

- 0.3*dP = 8.31*(-5) - P*dV

5. Since dP represents the approximate change in pressure, we can solve for it:

- 0.3*dP = -41.55 - 12*dP

- 12.3*dP = -41.55

- dP ≈ -41.55 / 12.3

6. Evaluating the expression, we find that the approximate change in pressure is approximately -3.38 kilopascals.

Therefore, when the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K, the pressure is estimated to decrease by approximately 3.38 kilopascals.