An ideal gas at $"155 kPa"$ and $25^@ "C"$ has an initial volume of $"1 L"$ . The pressure of the ideal gas increases to $"605 kPa"$ as the temperature is raised to $125^@ "C"$ . What is the new volume?

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An ideal gas at $"155 kPa"$ and $25^@ "C"$ has an initial volume of $"1 L"$ . The pressure of the ideal gas increases to $"605 kPa"$ as the temperature is raised to $125^@ "C"$ . What is the new volume?

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Answer 1 · 2018-03-05T02:38:20

My favorite combined gas law formula
$V_2= 0.342 L$

Explanation:

List givens, we cannot use temperature in Celsius, so convert to K, with K= C+273:
$V_1= 1 L$
$T_1=298 K$
$P_1= 155 KPa$
$V_2=?$
$P_2=605 KPa$
$T_2= 398 K$

Now Solve for $V_2$
By rearranging this formula: $(V_1P_1)/T_1=(V_2P_2)/T_2$
$V_2= (V_1P_1T_2)/(T_1P_2)$

Now just plug in numbers
$V_2= ((1 L)(155 KPa)(398 K))/((298 K)(605 KPa))$
$V_2= 0.342 L$

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An ideal gas at $"155 kPa"$ and $25^@ "C"$ has an initial volume of $"1 L"$ . The pressure of the ideal gas increases to $"605 kPa"$ as the temperature is raised to $125^@ "C"$ . What is the new volume?

1 Answer

Explanation:

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An ideal gas at "155 kPa""155 kPa" and 25^@ "C"25^@ "C" has an initial volume of "1 L""1 L". The pressure of the ideal gas increases to "605 kPa""605 kPa" as the temperature is raised to 125^@ "C"125^@ "C". What is the new volume?

1 Answer

Explanation:

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An ideal gas at $"155 kPa"$ and $25^@ "C"$ has an initial volume of $"1 L"$ . The pressure of the ideal gas increases to $"605 kPa"$ as the temperature is raised to $125^@ "C"$ . What is the new volume?