A container with a volume of #48# #L# contains a gas with a temperature of #140# #K#. If the temperature of the gas changes to #90# #K# without any change in pressure, what must the container's new volume be?

Answer 1

Using the Combined Gas Law, #(P_1V_1)/T_1=(P_2V_2)/T_2#, but leaving out pressure because it remains constant, #V_1/T_1=V_2/T_2 to V_2=(V_1T_2)/T_1=(48*90)/140=30.9# #L#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Aeschliman

Using the combined gas law equation, we have:

(P_1V_1/T_1 = P_2V_2/T_2)

Since the pressure remains constant, we can cancel it out, leaving us with:

(V_1/T_1 = V_2/T_2)

Substituting the given values:

(V_1 = 48 , \text{L}) (T_1 = 140 , \text{K}) (T_2 = 90 , \text{K})

(48/140 = V_2/90)

Solving for (V_2), we get:

(V_2 = (48 \times 90)/140)

(V_2 = 30.857 , \text{L})

Therefore, the container's new volume must be approximately 30.857 liters.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.