Answer to Question #249208 in Mechanical Engineering for russman

Question

A quantity of gas has an initial volume of 0.06 m³ and a temperature of 15 ⁰C. It is expanded to a volume of O. 12 m³ while the pressure remains constant. Determine the final temperature of the gas.

Answer 1

Q249208

A quantity of gas has an initial volume of 0.06 m³ and a temperature of 15 ⁰C. It is expanded to a volume of O. 12 m³ while the pressure remains constant. Determine the final temperature of the gas.

Solution :

In this question, the initial volume and temperature of the gas are given to us and we have to find the

the temperature of the gas when the volume is expanded to 0.12 m³.

Charle's law is the one that relates volume and temperature at constant pressure.

For two different sets of the condition, the law can be stated as

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

We are given the temperature in celsius. When using temperatures in the formula convert it to Kelvin.

The information given to us is

V₁ = 0.06 m³.

T₁ = 15 ⁰C = 15 + 273.15 = 288.15 K.

V₂ = 0.12 m³.

T₂ = unknown,

plug all this information in the formula we have

$$\frac{0.06 \ m^3 }{288.15 \ K } = \frac{0.12 \ m^3 }{T_2}$$

Arranging this equation for T₂, we have

$$T_2 = \frac{0.12 \ m^3 \ * \ 288.15 \ K }{0.06 \ m^3}$$

$$T_2 = \frac{34.578 \ K }{0.06 }$$

$$T_2 = 576.3 \ K$$

since the temperature given in the question was in celsius, so we will have to submit our final temperature in Celsius scale.

$$T_2 = 576.3 K = 576.3 - 273.15 = 303.15 \ ^0 C$$

In the question, the quantity with the least significant figures is 0.06 m³. It is in 1 significant figure, so our final answer must also be in 1 significant figure,

In the correct significant figure, the answer is 300 ^oC.

Hence the final temperature of the gas would be 300 ^oC.