Question

A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas by heat.

a) What is the work done on the gas?
b) What is the change in its internal energy?

Answer 1

The work done on the gas is 2.979 Joules. The change in internal energy of the gas is 114.021 Joules.

Step-by-step explanation:

In this question, we are given the initial and final volumes of a gas, as well as the constant external pressure and the heat transferred to the gas. To find the work done on the gas, we can use the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. Plugging in the values, we have W = (0.995 atm)(3.50 L - 0.200 L) = 2.979 J. Therefore, the work done on the gas is 2.979 Joules.

To find the change in internal energy of the gas, we can use the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat transferred to the gas, and W is the work done on the gas. Plugging in the values, we have ΔU = 117 J - 2.979 J = 114.021 J. Therefore, the change in internal energy of the gas is 114.021 Joules.

Answer 2

a)W= - 720 J

b)ΔU= 330 J

Step-by-step explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390 J

ΔU= 330 J