Physics Thermodynamics
Get insights from 156 questions on Physics Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Thermodynamics
- 156
Questions
- 0
Discussions
- 0
Active Users
- 1
Followers
Have a question about Physics Thermodynamics? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –QJ, during one cycle achieving and efficiency of 50.0%. The value of Q is:
A heat engine is involved with exchange of heat of 1915 J, –40 J, +125 J and –QJ, during one cycle achieving and efficiency of 50.0%. The value of Q is:
Efficiency η = Q_net/Q? = (1915-40+125-Q)/ (1915+125) = 0.5.
(2000-Q)/2040=0.5 ⇒ Q=980J.
Upvote 
A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
(f? n? RT? )/2 + (f? n? RT? )/2 = 3/2 RT + 5/2 × 3RT = 15
Upvote An engine takes in 5 moles of air at 20°C and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is:
An engine takes in 5 moles of air at 20°C and 1 atm, and compresses it adiabatically to 1/10th of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be X kJ. The value of X to the nearest integer is:
T? V? ¹ = T? V? ¹
T? = T? (V? /V? )? ¹ = T? (10)? /? ¹ = T? (10)²/?
ΔV = 5/2 nR; 5/2 × 5 × 3102/5-1
625/6 × 1.5 × 293 = 461440 ≈ 46ks
Upvote Initially a gas of diatomic molecules is contained in a cylinder of volume V₁ at a pressure P₁ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V₁, is given by P₂. The ratio P₂/P₁ is
Initially a gas of diatomic molecules is contained in a cylinder of volume V₁ at a pressure P₁ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V₁, is given by P₂. The ratio P₂/P₁ is
n' = 0.75n? +0.5n? = 1.25n? moles. P? /P? =5.
Upvote Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ (=C?/C?) are given, respectively, by:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ (=C?/C?) are given, respectively, by:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
f=5
∴ U = (5/2)RT
And γ = 1+2/f = 1+2/5=7/5
Upvote In adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be times the initial pressure. The value of is:
In adiabatic process, the density of a diatomic gas becomes 32 times its initial value. The final pressure of the gas is found to be times the initial pressure. The value of is:
In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128
Upvote