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Ivenika [448]
3 years ago
5

A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr

essure of
0.987 atm and the volume changes to 144mL?
Be sure to show all of your work, define variables, show equation and substitute variables. Report you answer in Celcius.
Chemistry
1 answer:
sertanlavr [38]3 years ago
4 0

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

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A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
sertanlavr [38]

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

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