## A sealed tank contains 30 moles of an ideal gas at an initial temperature of 270 K. The pressure of the gas is increased until the final pre

Question

A sealed tank contains 30 moles of an ideal gas at an initial temperature of 270 K. The pressure of the gas is increased until the final pressure equals 1.4 times the initial pressure. The heat capacity at constant pressure of the gas is 32.0 J(mol*K) What is the change in the internal energy of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).

130 kJ

77 kJ

-23 kJ

100 kJ

-50 kJ

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1 year 2021-08-21T07:12:58+00:00 1 Answers 22 views 0

40.4 kJ

Explanation:

The gas in this problem is in a sealed tank: this means that its volume is constant, so we can use the pressure law, which states that for an ideal gas kept at constant volume, the pressure is proportional to the temperature of the gas.

Mathematically:

$$\frac{p_1}{T_1}=\frac{p_2}{T_2}$$

where, in this problem:

$$p_1$$ is the initial pressure of the gas

$$p_2=1.4p_1$$ is the final pressure

$$T_1=270 K$$ is the initial temperature

$$T_2$$ is the final temperature

Solving for T2,

$$T_2=\frac{p_2 T_1}{p_1}=\frac{(1.4 p_1)(270)}{p_1}=378 K$$

Now we can find the change in internal energy of the gas, which is given by:

$$\Delta U=\frac{3}{2}nR(T_2-T_1)$$

where:

n = 30 mol is the number of moles

R = 8.314 J/(mol • K) is the gas constant

And substituting the values of the initial and final temperatures, we get:

$$\Delta U=\frac{3}{2}(30)(8.314)(378-270)=40406 J = 40.4 kJ$$