I have the following question as howework, I think I don't have enough information to solve it.

Given are $p_1=311$ kN/m$^2$, $T_1=333$ K, $T_2=853$ K.

Asked are $p_2$, ratio $V_1/V_2$, and $R$ if $w=372,32$ kJ/kg

How do I go about it? I can use the work equation to get a ratio between $k$ and $R$, but if I plug that into another equation I always remain with two unknowns.

I can't ask on Physics-meta, I'm sorry. I would really appreciate if someone showed me the way

  • $\begingroup$ What is R supposed to represent? $\endgroup$ – Chet Miller Jun 2 '18 at 19:58
  • $\begingroup$ The specific gas constant $\endgroup$ – Lazarus Jaeger Jun 2 '18 at 19:59
  • $\begingroup$ What is the exact statement of the problem? $\endgroup$ – Chet Miller Jun 2 '18 at 21:29
  • $\begingroup$ It's in dutch, but basically in an adiabatic transformation the temperature change is given, the starting pressure and the work done during that process. Asked it all the rest (end-pressure, change of volume etc ) $\endgroup$ – Lazarus Jaeger Jun 2 '18 at 22:06
  • $\begingroup$ Does it tell what gas it is? $\endgroup$ – Chet Miller Jun 2 '18 at 22:18

Although you’ve probably have already found the solution to the problem, here is one approach (though not necessarily the most efficient). It assumes (1) an ideal gas, (2) a closed (constant mass) system (such as a gas in a cylinder fitted with a piston and (3) an adiabatic and reversible (isentropic) process. The work, heat, internal energy, specific heats, and volumes are all lower case to indicate they are per unit mass (specific values).

We start with the work done in the reversible adiabatic (isentropic) process, given by:

(1) w = R(T1 – T2)/ (1-k)

Since the initial and final temperatures are given, we have two unknowns, R (specific gas constant, which you need to determine) and k.

For an ideal gas,

(2) k = c p / c v

where c v , the specific heat at constant volume, can be calculated as follows.

From first law:

Δu = q – w

q=0 (adiabatic process)

Δu = – w

w is positive if work is done by the gas (expansion) and negative if work is done on the gas (compression). Since the adiabatic processes increases the temperature (T2 > T1) the process must be compression, thus w is negative.

w = - 372.32 kJ/kg

Δu = - (- 372.32 kJ/kg) = +372.32 kJ/kg

But also, for an ideal gas in a closed system, Δu depends only on the temperature change and is given by the following that applies to any process (not just constant volume)

Δu = cv ΔT = cv (T2 – T1), then

(3) cv = +372.32 /(T2 – T1)

Next we introduce the relationship between specific heats and the specific gas constant for an ideal gas:

cp – cv = R


(4) cp = cv + R

recall (2)

k = c p / c v

and rewrite

(5) cp = k cv

From (4) and (5)

k cv = cv + R

(6) k = (cv + R)/ cv

substitute eq (3) for cv in eq (6), we then have k in terms of R, which can be substituted for k in equation (1) to solve for R. After solving R putting the value of R into equation 1 can solve for k.

That leaves v1/v2 and p2 to be determined. We get these from the equation for a reversible adiabatic process, and the general gas equation for an ideal gas.

Process equation (reversible adiabatic- isentropic):

Pvk = constant

P1v1k = P2v2k

(7) v1/v2 = (P2/P1)1/k

General gas equation:

P1v1/T1 = P2v2/T2

(8) v1/v2 = P2T1/P1T2

Since k has been determined, we have two equations ( (7) and (8) ) and two unknowns v1/v2 and p2 which can then be solved.

  • $\begingroup$ BTW, can anyone advise the best way to draft comments with respect to equations? I use Word. I tried using the formula approach as well as simple superscripts and subscripts, but none seems to come out right. For example v1k should be volume 1 raised to the k power. $\endgroup$ – Bob D Jul 15 '18 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.