# A cylinder of fixed capacity, 66.4 liters, contains helium gas at standard temperature and...

## Question:

A cylinder of fixed capacity, 66.4 liters, contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 18.0 degrees C? (R=8.31Jmol{eq}^{-1} {/eq}K{eq}^{-1} {/eq}).

## Heat Transfer:

The total heat transferred from one substance to another is determined from the change in the temperature of one substance, assuming that the system is thermally isolated from the environment. We compute for the total heat as {eq}\displaystyle q = mc\Delta T {/eq}, wherein *m* is the mass, *c* is the specific heat, and {eq}\displaystyle \Delta T {/eq} is the change in the temperature of the substance.

## Answer and Explanation:

Determine the total heat, *q*, by applying the equation, {eq}\displaystyle q = mc\Delta T {/eq}. We have the mass, *m*, which can be derived from the product of the number of moles, {eq}\displaystyle n = \frac{PV}{RT} {/eq} from the ideal gas equation, and the molar mass of He, {eq}\displaystyle MW = 4.00\ g/mol {/eq}, or {eq}\displaystyle m = n\times MW {/eq}. We use the values:

- {eq}\displaystyle P = 101325\ Pa {/eq}

- {eq}\displaystyle V = 66.4\ L = 0.0664 m^3 {/eq}

- {eq}\displaystyle T = 298\ K {/eq}

- {eq}\displaystyle R = 8.31\ J/mol\ K {/eq}

Moreover, we use the specific heat of He as {eq}\displaystyle c = 5.193 J/g K {/eq} and the change in temperature as {eq}\displaystyle \Delta T = 18.0^\circ C = 18.0\ K {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ &= \left( n\times MW \right) c\Delta T\\ &=\left( \frac{PV}{RT}\times MW \right) c\Delta T\\ &= \left( \frac{101325\ Pa\times 0.0664 m^3}{8.31\ J/mol\ K\times 298\ K}\times 4.00\ g/mol \right) (5.193 J/g K)(18.0\ K)\\ &\approx\boxed{\rm 1016\ J} \end{align} {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12