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What is the temperature of 0.49 mol of gas at a pressure of 1.2 ATM and a volume of 10.5 L?
Wiki User
∙ 9y ago
I am going to show you a way of doing this.
As 1 atmosphere is equivalent to 1 kPa (101.325 kPa) we can multiply 101.325 * 1.2 to get the pressure in kilopascals.
1.2 * 101.325 = 121.59 kPa
We can use the ideal gas law and substitute to find the rest. Remember to use R = 8.314 k/j mol ^-1 for kilopascals and R = 0.0821 for atmospheres
pV=nRT
121.59 * 10.5 = 0.49* 8.314 * T
1276.7/4.07386 = T
313.39* C = T
But as Temperature is in Kelvin add on 278.15.
Therefore, T = 591.54 K
Wiki User
∙ 9y ago
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