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Quantity of ideal gas whose ratio of principal molar heat capacities is 5:3 has a temperature 300K, volume 0.064m^3 and pressure 243kpa.it reversible isothermal of volume 0.064.calculate temperature ?

Apr 18

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Apr 18

To solve this problem, we can use the ideal gas law and the relationship between the principal molar heat capacities.

Given:

- Temperature (T1) = 300 K

- Volume (V1) = 0.064 m^3

- Pressure (P1) = 243 kPa

- Ratio of principal molar heat capacities (γ) = 5:3

We know that for an ideal gas undergoing an isothermal process, the pressure and volume are inversely proportional. Therefore, we can use the equation:

P1V1 = P2V2

Substitute the given values into the equation:

243 kPa * 0.064 m^3 = P2 * 0.064 m^3

P2 = 243 kPa

Since the process is reversible and isothermal, the temperature remains constant. Therefore, the temperature (T2) will also be 300 K.

Therefore, the temperature of the gas remains at 300 K during the reversible isothermal process.

Given:

- Temperature (T1) = 300 K

- Volume (V1) = 0.064 m^3

- Pressure (P1) = 243 kPa

- Ratio of principal molar heat capacities (γ) = 5:3

We know that for an ideal gas undergoing an isothermal process, the pressure and volume are inversely proportional. Therefore, we can use the equation:

P1V1 = P2V2

Substitute the given values into the equation:

243 kPa * 0.064 m^3 = P2 * 0.064 m^3

P2 = 243 kPa

Since the process is reversible and isothermal, the temperature remains constant. Therefore, the temperature (T2) will also be 300 K.

Therefore, the temperature of the gas remains at 300 K during the reversible isothermal process.

Google Gemini AI

Apr 18

Given:

$$\frac{C_p}{C_v}=\frac{5}{3}$$

T_1=300K

V_1=0.064m^3

P_1=243kPa

V_2=0.32m^3

We know that,

$$C_p - C_v = R$$

$$\frac{C_p}{C_v} = \frac{C_p}{C_p - R}$$

$$\frac{5}{3} = \frac{C_p}{C_p - R}$$

$$5(C_p - R) = 3C_p$$

$$5C_p - 5R = 3C_p$$

$$5C_p - 3C_p = 5R$$

$$2C_p = 5R$$

$$C_p = \frac{5R}{2}$$

Now,

$$PV = nRT$$

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

$$T_2 = \frac{P_2V_2T_1}{P_1V_1}$$

$$T_2 = \frac{(243\times10^3)(0.32)(300)}{(243)(0.064)}$$

$$T_2 =1200K$$

Therefore, the temperature is 1200K.

$$\frac{C_p}{C_v}=\frac{5}{3}$$

T_1=300K

V_1=0.064m^3

P_1=243kPa

V_2=0.32m^3

We know that,

$$C_p - C_v = R$$

$$\frac{C_p}{C_v} = \frac{C_p}{C_p - R}$$

$$\frac{5}{3} = \frac{C_p}{C_p - R}$$

$$5(C_p - R) = 3C_p$$

$$5C_p - 5R = 3C_p$$

$$5C_p - 3C_p = 5R$$

$$2C_p = 5R$$

$$C_p = \frac{5R}{2}$$

Now,

$$PV = nRT$$

$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$$

$$T_2 = \frac{P_2V_2T_1}{P_1V_1}$$

$$T_2 = \frac{(243\times10^3)(0.32)(300)}{(243)(0.064)}$$

$$T_2 =1200K$$

Therefore, the temperature is 1200K.